
Class Tfy^yO , 

Rnnk ~H S 



COPYRIGHT DEPOSm 



GRAPHICS 



AND 



STRUCTURAL DESIGN 



H. D. HESS, M.E. 



BY 

' J 

Professor of Machine Design, Sibley College, Cornell University. Formerly Designer 

and Computer for the Mechanical Department of the Pencoyd Iron 

Works and the American Bridge Co. Member of the 

American Society of Mechanical Engineers. 



SECOND EDITION 

FIRST THOUSAND 



NEW YORK 

JOHN WILEY & SONS, Inc. 

London: CHAPMAN & HALL, Limited 

1915 



vO>\S 



Copyright, i9i5"« 

BY 

BL D. HESS 



Staitbope jpress 

F. H.GILSON COMPANY 
BOSTON. U.S.A 



9h- ^_ ^^ 



nri -9 1915 

CI.A411936 



PREFACE 



This text is intended for the author's classes in General 
Engineering Design in Sibley College, Cornell University. The 
treatment of the subject has been kept as general as possible and 
is that used for his students during the past four or five years. 
The problems chosen for discussion are those on the border line 
between Civil and Mechanical Engineering. To the general 
designer knowledge of this subject is indispensable while acquaint- 
ance with the methods used in determining the stresses in, and 
the subsequent designing of, structures is of the greatest benefit 
to others in all designing for strength whether of machines or 
structures. 

Although the book is called "Graphics and Structural Design" 
the determination of the stresses has not been confined to graphi- 
cal means; the other usual methods have been included. 

The design of a plate girder may seem out of keeping with 
the purpose of the book. The author formerly used a crane 
runway girder instead of the railway bridge, but decided that 
the railway girders made the more comprehensive and better 
problem. The design of the latter permitted the use of the 
moment table and acquainted the student with the usual method 
of treating locomotive and train loads. 

Incidentally, one who could design a railway plate girder was 
well prepared to design a runway girder, although the reverse 
was not so generally true. 

The author's practice is to use a set of problems paralleling the 
drawing-room work, to assign a number of these problems each 
week to the students upon which they recite. Although this 
reduces slightly the time spent over the drawing boards the work 



IV PREFACE 

accomplished by the students is not apparently diminished and 
the work seems to be generally better under stood. 

Although intended primarily for a textbook, it is hoped that 
this may prove a satisfactory reference book for designers whose 
work is not too highly specialized. 

The author desires to acknowledge his indebtedness to many 
manufacturers, all of whom have been most generous and among 
whom are American Bridge Co., Pennsylvania Steel Co., Cam- 
bria Steel Co., Bethlehem Steel Co., McClintock-Marshall Con- 
struction Co., Jeffrey Mfg. Co., Best & Co., and Bucyrus Co. 

Among the periodicals he is especially indebted to the 
Engineering News. 

The section on Specifications has been drawn largely from the 
specifications of the American Railway Engineering and Main- 
tenance of Way Association, the specifications of Mr. C. C. 
Schneider in the Transactions of the American Society of Civil 
Engineers, and the specifications of the American Bridge Co. 

H. D. HESS. 

Ithaca, N. Y. 
May, 1913° 



PREFACE TO SECOND EDITION 



A number of changes have been made, including a more ex- 
tended treatment of steel mill-buildings and of reinforced con- 
crete, modifications and explanations of the specifications, and 
other changes due to the advances made in the subject. 

H. D.H. 

Ithaca, N. Y. 
August, 1915. 



CONTENTS 



CHAPTER I. 

MATERIALS. 

Cast Iron — Steel — Steel Castings — Physical Properties of Metals — Proper- 
ties of Timbers — Timber Beams — Timber Columns — Bricks — Table of 
Bending Moments, Deflections, Etc. — Table of Properties of Sections — 
Strength of Flat Plates — Properties of Sections of Rolled Structural Steel. 1-19 

CHAPTER II. 

GRAPHICS. 

Force — Magnitude, Direction and Line of Action of a Force — Equilibrium — 
Couple — Resultant — Equilibrant — Force Triangle — Force Polygon — 
Components — Composition of Forces — Resolution of Forces — Graphic 
Moments — Moments of Parallel Forces — Uses of Force and Equilibrium 
Polygons — Graphical Determination of Deflections — Graphical Analyses of 
Restrained and Continuous Beams 20-34 

CHAPTER III. 

STRESSES IN STRUCTURES. 

Tension or Compression in a Member — Stresses due to Moving Loads — Wind- 
load Stresses — Maximum Bending due to Moving Loads — Maximum Live- 
load Shears 35-5 1 

CHAPTER IV. 
ALGEBRAIC DETERMINATION OF STRESSES. 

Method of Moments — Method of Coefficients 52-60 

CHAPTER V. 

INFLUENCE DIAGRAMS. 

Position of Loading for Maximum Moments — Position of Loading for Maximum 
Shear — Position of Loading for Maximum Floor-beam Reaction — Moment 
Table for Cooper's Loading, E-60 61-75 



VI CONTENTS 

CHAPTER VI. 

TENSION PIECES, COMPRESSION PIECES AND BEAMS. 

Tension Pieces — Compression Pieces — Beams — Vertical Shear — Bending 
Moment — Standard Framing — Riveting — Shearing and Bearing Value of 
Rivets — Rules for Spacing Rivets . 76-90 

CHAPTER VII. 
COLUMNS. 

Column Formulae — Combined Stresses — Long Beams Unsupported Laterally. 

91-102 

CHAPTER VIII. 

GIRDERS FOR CONVEYORS. 

Dead and Live Loading — Wing Loading — Determination of Stresses — Selection 
of Members 103-1 1 1 

CHAPTER IX. 

TRUSSES, BENTS AND TOWERS. 

Trusses and Bents for carrying Pipes — Towers for Electrical Transmission 
Lines 112-119 

CHAPTER X. 
DESIGN OF STEEL MILL BUILDINGS. 

Loading — Bracing — Determination of Stresses — Selection of Members. 120-143 

CHAPTER XL 

DESIGN OF A PLATE-GIRDER RAILWAY BRIDGE. 

Loading — Bending Moments — Vertical Shears — Flange Area and Selection 
of Sections — Lengths of Flange Plates — Spacing of Flange Rivets — Wind 
Bracing — End Diagonals — Web Splice — Flange Splice 144-164 

CHAPTER XII. 

CRANE FRAMES. 

Design of a Frame for an Underbraced Jib Crane — Design of a Frame for a 
Top-braced Jib Crane 165-176 



CONTENTS Vll 

CHAPTER XIII. 

GIRDERS FOR OVERHEAD ELECTRIC TRAVELING CRANES. 

Box Girders — Plate Girders with Upper Flange Stiffened by a Channel — Bridge 
Girders with Horizontal Stiffening Girders 177-194 

CHAPTER XIV. 

REINFORCED CONCRETE. 

Cement — Sand — Stone and Gravel — Physical Properties of Concrete — Unit 
Working Stresses — Rectangular Beams — Approximate Formulas — T Beams — 
Beams with Double Reinforcing — Web Stresses — Bond Stresses — Lengths 
of Reinforcing Rods — Bent Bars — Web Reinforcements — Columns. 195-232 

CHAPTER XV. 

FOUNDATIONS. 

Materials — Allowable Pressure on Soils — Machine Foundations — Building 
Foundations — Piles 233-246 

CHAPTER XVI. 

CHIMNEYS. 

Kern of a Section — Design of a Brick Chimney — Design of a Self-sustaining 
Steel Chimney — Design of a Reinforced-concrete Chimney . .' 247-281 

CHAPTER XVII. 

RETAINING WALLS. 

General Theory of Retaining Walls — Angles of Repose of Materials — Design 
of Concrete Retaining Walls — Design of Reinforced-concrete Retaining 
Walls 282-297 



CHAPTER XVIII. 

BINS. 

Determination of Pressures on the Sides and Bottoms of Bins — Stresses in Bins — 
Graphical Determination of the Forces acting on Bins — Hopper Bins — 
Suspension Bunkers — Bin Design 298-322 



Vlll CONTENTS 

CHAPTER XIX. 

SHOP FLOORS. 

Cement Concrete — Wooden Blocks — Asphalt — Brick — Wooden — Upper 
Floors, Wooden — Steel — Brick Arch — Reinforced Concrete — Iron . . 323-331 

CHAPTER XX. 

WALLS AND ROOFS. 

Wooden Sides — Corrugated Steel Sides — Walls, Brick — Stone — Solid Con- 
crete — Hollow Concrete — Reinforced Concrete — Windows — Roof Cover- 
ings, Corrugated Steel — Slate — Clay Tiles — Concrete — Slag or Gravel. 

332-352 

CHAPTER XXI. 

SPECIFICATIONS. 

General, Materials — Workmanship — Painting — Inspection — Testing — Speci- 
fications for Steel Mill Buildings — Specifications for a Deck Plate-girder 
Railway Bridge — Specifications for Reinforced Concrete 353 - 388 

CHAPTER XXII. 

Problems 389-424 



GRAPHICS 
AND STRUCTURAL DESIGN 



CHAPTER I 
MATERIALS 

The principal materials used in the structures considered in 
this book are iron castings, hard, medium and soft steels, steel 
castings, timber (oak, hemlock, long-leaf Southern and yellow 
pine, spruce and white pine) and concrete, cement, slag and 
stone. 

Cast Iron is made by remelting pig iron or pig iron and cast- 
iron scrap in a cupola. The quality of the product is a varying 
one, as the entire charge in the cupola is never melted at any 
one time. The quality, therefore, depends upon the time at 
which it was run off during the heat. 

For machining, cast iron should be soft, and should have an 
ultimate strength in tension of from 16,000 to 20,000 lbs. per 
sq. in. Although cast iron has no well-defined elastic limit, it 
may be assumed for practical purposes at 8000 lbs. per sq. in. 
Cast iron is exceedingly strong in direct compression; when no 
bending is introduced its ultimate strength in compression should 
reach 90,000 to 100,000 lbs. per sq. in. The resilience of cast 
iron being very low indicates little ability to resist shock. 

Steel is made by refining pig iron in the Bessemer converter, 
also by refining pig iron or pig iron and steel scrap in the open- 
hearth furnace. Small quantities of special steels are also being 
made in electric furnaces. The refining processes remove the im- 
purities and the carbon, the steel being afterwards recarbonized. 
The ordinary merchant and structural steels are soft and medium 



GRAPHICS AND STRUCTURAL DESIGN 



steels, the carbon contents ranging from 0.08 to 0.30 of 1 per 
cent. The ultimate strength of steel varies with the percentage 
of carbon from 48,000 lbs. per sq. in. for very soft or rivet steel 
to 70,000 lbs. per sq. in. for medium steel. Steel, having great 
resilience, is peculiarly adapted to resisting shock. 

Steel Castings. — This steel is also made in open-hearth 
furnaces but may be made in small converters. The steel is 
poured into molds as is done for iron castings. To obtain 
fluidity it is necessary that steel for castings have a much higher 
temperature than is required for cast iron. The mold frames 
must, therefore, be much more substantial and the molds must 
be thoroughly dried before the metal is poured into them. 

Considerable contraction, with the attendant internal stresses, 
results from the exceedingly high temperature and necessitates 
the annealing of steel castings. The ultimate strength depends 
upon the carbon and should be 50,000 or more pounds per 
square inch. Steel castings have high resilience and are, there- 
fore, superior to iron castings. However, both the material 
and the machining of steel castings cost more than do those 
of iron castings. 

TABLE OF PHYSICAL PROPERTIES OF METALS 





Modulus of elasticity. 


Ultimate strength. 


Elastic strength. 


Material. 


Tension 
compres- 
sion. 


Torsion. 


Tension. 


Shear. 


Tension. 


Shear. 




i 10,700,000 
( 15,000,000 

{:::::::: 

28,000*000 

30,000,000 
30,000,000 
31,000,000 

30,600,000 


4,000,000 
6,000,000 

11,000,000 

11,800,000 
11,800,000 
12,100,000 

11,800,000 


16,000 

20,000 

30,000 

40,000 

{ 47,ooo 

( 57,000 

60,000 

70,000 

100,000 

| 50,000 

\ 100,000 


16,000 ) 
20,000 j 

35.ooo 
43,ooo 
45,000 
52,000 

30,000 
60,000 


8,000 

40,000 
45,000 
75,000 


8.000 


Cast iron (air furnace) 

Wrought iron 








Steel 0.15 carbon 




Steel 0.25 carbon 




Crucible steel (high carbon) . . . 









Note. — The ultimate compressive strength of cast iron is 90,000 to 
100,000 lbs. per sq. in. Its elastic strength in compression may be taken 
at 25,000 lbs. per sq. in. The ultimate compressive strengths of the other 
materials will approximate their ultimate tensile strengths. 



MATERIALS 3 

Working Fiber Stresses. — In the structures hereinafter de- 
scribed the usual working fiber stresses will be given in each 
problem. Ordinarily, in structures liable to little or no shock 
or vibration, where the stresses are fully determined, the 
maximum fiber stress on mild steel will range from 16,000 to 
20,000 lbs. per sq. in. Structures, such as crane frames, liable 
to some shock will have the maximum stress on mild steel 
reduced to 11,000 or 12,000 lbs. per sq. in. In the case of high- 
way bridges and similar structures an addition of 25 per cent is 
added to the live-load stresses to allow for impact, and a unit 
stress of 15,000 lbs. per sq. in. is allowed on soft steel, while 
a unit stress of 17,000 lbs. per sq. in. is allowed on medium 
steel. In the case of railway bridges 15,000 and 17,000 lbs. 
per sq. in. are used, but the impact allowance is made by a 
formula similar to that used in the design of the railway plate 
girder in Chapter XI. 

Properties of Timber 

Wooden beams are designed similarly to metal ones. Being 
liable to fail by horizontal shearing they should be examined for 
this. One-fifth the ultimate shearing resistance given in the tables 
may be taken as the working shearing strength (lengthwise). 

The formula for bending is 

M=f~- 

e 

M = bending moment in inch pounds. 

/ = moment of inertia in inches 4 . 

e = distance from the neutral axis to the extreme fibers in 

inches. 
/ = working fiber stress, in flexure, pounds per square inch. 
For rectangular timber beams this becomes 

6 
b = width of the beam in inches. 
d = depth of the beam in inches. 



GRAPHICS AND STRUCTURAL DESIGN 



The maximum longitudinal unit shearing stress is /, = 

where /, = fiber stress in shear, pounds per square inch. 
R = end reaction in pounds. 



3_*. 
2bd' 



Timber Columns 
The following formula is suggested by the United States 
Government reports on timber. 



700+15C + C 2 



/< 



ultimate compressive strength of the column in pounds 
per square inch. 
F c = ultimate crushing strength of short timber column in 
pounds per square inch. 
I 
C = d' 
where 

I = length of the column in inches. 
d = least diameter in inches. 

The factor of safety should vary from 5 for 18 per cent moisture 
when used in the open to 3 J for 10 per cent or less of moisture 
when used in heated buildings. 



Name of wood. 


Pounds per 

square inch 

Fc- 


White oak, Southern long-leaf pine 

Short-leaf yellow, pine (Georgia) 

Hemlock, chestnut and spruce 


5000 
4500 
4000 
3500 


White pine and cedar 





The properties of concrete will be treated quite fully under 
Reinforced Concrete; see Chapter XIV. 

Bricks. — Bricks will vary in size and properties according 
to the locality in which they are made. Common bricks may 
generally be assumed about 8^ ins. X 4 ins. X 2 J ins., while face 
bricks will run 8| ins. X 4! ins. X 2J ins. Common bricks will 



MATERIALS 



•}OOj oiqno jad 
spunod '^qSia^w 



a» o 
E-i'55 



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'dm 
O cS 

SH3 












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8888 

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OOO 



888 

<N O On 



o o o 
o o o 

On I s * t^ 



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O-nO 



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o ^ o 

too "3- 



8888888 

00 CO H On m hh On 



o o o o o 
o o o o o 

o no <r> o o 



0000 NNt^N CNCOHOOO 



o o o o 

o o o o 

o_ o o_ o_ 

6 6 6 6 

■* M ^t-OO 

vO O i -1 *0 



o o 
o o 

OOO 




oo 



OOO 
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O M <N 



o 


O O 

r^ co 
CO to 


to O O 

r-» ^f to 


O • 
O ■ 

VO • 


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. to • 


'O 


H H 


CM ^ cs 


to 


. CO . 



o o o o 
o o o o 

to ^ O <* 



to o o o o 
co r^ O O O 

00 f-00 -^too 



OOO ■ o 
no O to • o 

IN N H 



o o o o o o 
o o o o o o 

CM CONNIOTJ- 



o o o o o 
o o o o o 

O r^ ^t O co 



t^. to to to tooo to oo oo x^-no t-- 



o o o o o 
o o o o o 

to O f^NO O 



O O O O O 

o o o o o 
o o o o o 






£ -d o ^ -° ^ Z • 

\Z i-< V m r-t D <—< 



3 w 



*c^ 



C/D^ 



<D 3J3 



03 * If^U a; o S o o ? p" 



,c! 13 a; 



C C^ 



t- J-l 



<OOQffiOft,dH PhPh c# c# 









c 

<1) 




a 


ft 


ft 


r/3 


a> 


Tt 


ft 


a 




3 




o 


4J 
4J 


ft 


aj 


rl 






m 


<u 


13 





1 crj 

<:fti 



6 GRAPHICS AND STRUCTURAL DESIGN 

weigh 125 lbs. per cu. ft., while soft bricks will average 100 lbs. 
per cu. ft. 

Good bricks should be sound, hard, of regular shape and size, 
and should have a uniform structure. When struck a sharp blow 
they should emit a ringing sound. They should not absorb more 
than t X q their weight of moisture. Their crushing strength 
should reach 7000 to 10,000 lbs. per sq. in. 

Fire Bricks. — These are bricks made of clay or of a mixture 
of clay and sand having ability to resist high temperatures. 
They are used mainly for building furnaces and lining flues and 
some chimneys. 



MATERIALS 



BENDING MOMENTS, DEFLECTIONS, ETC., FOR BEAMS OF 
UNIFORM SECTION 



Form of support and 
load. 





End reactions A , B. 
Bending moment M. 



B = P, 
M=Px, 

Mmax =Pl. 



A = B = 



M- 



M max = 



Px 

2 
PI 



Relation between 
load " P " and mo- 
ment of resistance. 



p _fW 
W.S. 



P = 4 



fW 



4/ 



Maximum deflection. 



P P 

EI 3 

3 Ex 



£1*48 



12 E-e 




A=^,B^ 



P = 



fWl 



-Mmax = 



Pec 1 
I 



W= 



Pec 1 
l.f 




A=B = P, 

Between A & B, 
M=P'C = const. 



P-«. 



f 



d = 



Pcci 



27 Ell 
X (c + 2 d) 



X\/3^(c+2Ci). 



d x 



= P_Pc 

1 £/ 8 V 

*f I 2 



8£ e 



_ P fc 3 . cH\ 

-Elll + Tj 




B = P, 
™ Px * 

2 I 

M Pl 

-Mmax — — • 



w- 



fw 



PI 



p_ p 

£7*8' 



4Ee 



A=B- 






M 



Px ( x\ 
"TV"!) 1 



Mmax = 



PI 



P = 



W-- 



SfW 



PI 



8/' 






a 48 £e 



GRAPHICS AND STRUCTURAL DESIGN 



BENDING MOMENTS, DEFLECTIONS, ETC., FOR BEAMS OF 
UNIFORM SECTION 



Form of support. 




"T 3 !^ 



r 



End reactions A , B. 
Bending moment. 



A = -P, B = - P, 



M 



-(-?) 



Mmas = O.I28 Pl. 



Relation between 
load " P " and mo- 
ment of resistance. 



P = 7794 



fw 



w = 



PI 



7.794/ 



Maximum deflection. 



d = 0.01304 



PI* 
EI 



\ + 


*^ ' 


■ 


, P^"* 


^1 


(A) 

! 


r 


> 


^d 


1 1 1 

- J 






n 



4 = B=£, 

2 



P-af, 



"-*(H*)' 

Mmax = -r ' 



w = 



£1 

6/' 



P/ 3 
60 £/ 



10 Ee 




4-|P. *-§P, 



Mr 



2 \4 I 
PI 



W = 



SfW 
I ' 



PZ 
8/" 



ttmax — 



P/ 3 



185 £/ 



^.i d r-*- 




A = \P, B = -^P, 
16 16 



Mmax = 



K J- *i 



111 

16 



16 JW 

3 / 

16 / 



-vl 



PI* 

5 482S/' 




4 = 5 



M = -7{6-J + p) 



Mmax = — 



P = 



IF 



12 fW 



PI 
12/ 



P/ 3 



384 £/ 



_L ZL 2 . 

32 £e 



bill* 



p = 



/w 



p/ 3 



Mmax = -7T- 



w = 



-I H 



Pl 

8/ 



192 EI 
P 
24 Ee 



-6= a >j 



IB- 



-Axif- 



^ 






*=3 -p(.+¥-i)- 



MATERIALS 



The bending moment is a maximum under load when x = - ■ 

4 



TUT Pl l a \ 



The letters used have the following significance and for con- 
venience they should be expressed in the units stated. 
A and B = end reactions, e = distance neutral axis to 

pounds. extreme fibers having 

P = load, pounds. fiber stress /, inches. 

/ = extreme fiber stress, E = modulus of elasticity, 

pounds per square inch. pounds per square inch. 

/ = span of beam in inches. c, C\ and x = portions of span, 

inches. 

W = resistance, = -• / = moment of inertia, inches 4 . 

e 

d = deflection, inches. 

Strength of Flat Plates 

Nomenclature: 

t = thickness of plate in inches. 
f m = maximum fiber stress in plate, pounds per square inch. 
E = modulus of elasticity in flexure, pounds per square inch. 

c factor according to Grashof or Bach. 

P = concentrated load in pounds, usually assumed as acting 
on a circle whose radius is r ins. 

q = uniform load acting on plate, pounds per square inch. 

r = radius of circular plate, inches. 

b = longer side of rectangular plate, inches. 

a = shorter side of rectangular plate, inches. 

b = 

a 
i. Circular plates, carrying uniform load q. 



IO 



GRAPHICS AND STRUCTURAL DESIGN 



MOMENTS OF INERTIA, RESISTANCES, CENTERS OF GRAVITY 

AND LEAST RADII OF GYRATION OF 

GEOMETRICAL SECTIONS 



Shape of 
section. 




I 



i-«— 5-»i 



Moment of inertia. 



12 



£ 4 -& 4 



bW 
36 



Resistance. 



6 



i ff 4 -6 4 
6 5 



M_ 2 
24' 



Distance base 

to center of 

gravity. 



Least radius of 
gyration. 



Lesser side 
3-46 



\l 



B 2 + W 



h 

4.24 






6& 2 +66&i+&i 2 

3 6( 2 h-&i) 



1 3^+61 
3 2&4-&J 







D 4 



20.38 



D 3 



10.19 



D)[ 



0.049 (£> 4 -d 4 ). 



0.098 



£> 4 -J 4 



iV(Z) 2 +^) 



-£5 



1 



0.11R*. 



W! = o.igiR\ 
^2=0.259 R 3 . 



0.424 R. 



0.07 i? 2 . 



-UAX 
6H- 




0.7854 &a 3 . 



o. 78546a 2 



MATERIALS II 

Cast-iron plate supported at outer circumference, c = 0.8 to 1.2. 
Steel plate supported at outer circumference, c = 0.75. 

Steel plate bolted or riveted at outer circumference, c = 0.50. 

2. Circular plates, carrying a central load P, applied on a circle 
of radius r Q . Plate supported at outer circumference. 

1.05 \ 2>r ) f 

For cast iron c = 1.5. 

3. Circular plates, central load similar to 2 but plate bolted 
or riveted at outer circumference. 



P, r 



/- = ^ lo gr 



Rectangular and Square Plates 

Rectangular and square plates supported at the outer edges 
and carrying a uniform load q. 
Rectangular, 

A 2 
2?" 



/» - c _ /2 x T 1 v 



Square, 



■ , ca z 



For cast iron c = 0.75 to 1.13. 

For steel c = 0.56 to 0.75 (max. 1.13). 



Structural Material 



The principal rolled sections used are I beams, channels, 
angles, plates, flats and rounds. Manufacturers' handbooks 
afford the best sources of information regarding these sections 
and in actual design they should be freely consulted. The 
following data are abridged from these books. 



12 



GRAPHICS AND STRUCTURAL DESIGN 



SHEARED PLATES 



Width in inches. 



20 
21 

22 

23 
24 

25 
26 

27 
28 
29 



30- 35 
36- 41 
42- 47 

48- 53 
54- 59 

60- 65 
66- 71 
72- 77 
78-83 
84- 89 

90- 95 

96-101 

102-107 

108-113 

114-119 

120-125 



Thickness in inches. 



1 

4 


A 


3 

8 


A 


2 


9 


5 

8 


11 

T6 


3 

4 


if 


2 15 
8 T6 


I 


ll 


li 



I* 



Length in inches. 



240 

240 
240 
240 
240 

240 
216 
216 
204 
204 

204 
180 
168 
168 
156 



320 
320 
320 
360 
360 

360 
360 
360 
360 
360 

360 
360 
340 
340 
340 

320 
360 
360 
360 
340 

320 
300 
260 
240 
200 

180 

120 



400 
400 
400 
400 
400 

400 
400 
400 
400 
400 

400 
400 
400 
400 
400 

400 
400 
400 
400 
400 

400 
350 
300 

275 
250 

230 
175 

I50 



500 
500 
500 
500 

500 

500 

500 
500 
500 
500 

500 

500 
500 
500 
500 

500 
500 
500 
500 
500 

500 
430 
400 
38o 
350 

330 

240 
200 
180 



500 
500 
500 
500 
500 

500 
525 
525 
525 
525 

525 
525 
500 
500 
500 

500 

500 
525 
525 
500 

500 

45° 
425 
400 

375 

34o 
250 
230 
180 



55o 
55o 

55o 
500 
500 

500 

525 
525 
525 
55o 

55o 

55o 
55o 
55o 
55o 

55o 
500 
525 
55o 
55o 

55o 
475 
45o 
420 

385 

35° 
275 
230 
200 
180 

120 



500 
500 
500 
500 

500 

500 
525 

525 
525 
55o 

550 

55o 
55o 
55o 
55o 

500 
500 
525 
55o 
55o 

500 

425 
400 

375 
35o 

35o 

275 
250 
220 
200 

150 



475 
475 
475 
55o 
55o 

55o 
55o 
55° 
55o 
55o 

55o 
55o 
500 
500 
500 

475 
55o 
55o 
55o 
500 

475 
425 
400 

375 
35o 

325 
275 
250 
225 
210 

150 



475 
475 
475 
55o 

55o 

55o 
55o 
55o 
55o 
55o 

55o 
55o 
500 
500 
500 

475 
55o 
55o 
55o 
500 

475 
425 
400 

375 
35o 

325 
275 
250 
225 
210 

180 



475 
475 
475 
55o 
55o 

55o 
550 
550 
550 
55o 

550 
55o 
500 
500 
500 

475 
55o 
55o 
55o 
500 

475 
410 

39o 
37o 
35o 

325 
275 
250 
225 
210 

180 



425 
425 
425 
500 

500 

500 
500 
500 
500 
500 

500 

500 
450 
450 
45o 

425 
500 
500 
500 
45o 

425 
375 
35o 
325 
300 

275 
240 
230 
220 
200 

175 



400 
400 
400 
500 
500 

500 
500 
500 
500 
475 

475 
475 
45o 
45o 
45o 

400 
500 

500 
475 
45o 

400 
34o 
320 
300 
280 

260 
240 
230 
220 
200 

175 



375 
375 
375 
450 
45o 

45o 
45o 
45o 
45o 
425 

425 
425 
400 
400 
400 

375 
45o 
45o 
425 
400 

375 
33o 
320 
300 
275 

260 
220 
210 
200 
180 

160 



360 
360 
360 
400 
400 

400 
400 
400 
400 
400 

400 
400 
380 
380 
380 

360 
400 
400 
400 
380 

360 
320 
300 
300 
275 

260 
220 
210 
200 
180 

160 



300 
300 
300 
400 
400 

400 
400 
400 
400 
35o 

35o 
35o 
330 
33° 
33o 

300 
400 
400 
35o 
33° 

300 
280 
260 
240 
230 

220 
200 
190 
180 
170 

160 



280 
280 
280 
400 
400 

400 
400 
400 
400 
35o 

35o 
35o 
300 
300 
300 

280 
400 
400 

35o 
300 

280 
260 

240 
220 
210 

200 
180 
170 
160 
150 

144 



280 
280 
280 
35o 
35o 

35o 
35° 
35o 
35o 
325 

325 
325 
300 
300 
300 

280 
35o 
35o 
325 
300 

280 
260 
240 
220 
210 

200 
180 
170 
160 
150 

144 



MATERIALS 
EDGED PLATES 



13 















Thickness 


in inches. 












Width in 


3 


1 


5 


3 


7 


1 


9 


5 


11 


3 


it 


7 


15 




inches. 


T6 


4 


T6 


8 


T6 


2 


T6 


8 


16 


4 


8 


16 


I 




Length in feet. 


4 


50 


50 


50 


5° 


50 


50 


40 


40 


30 


30 


30 


28 


28 


28 


5 


30 


42 


42 


42 


42 


40 


30 


30 


30 


30 


30 


30 


30 


30 


6 


30 


42 


42 


42 


42 


40 


35 


30 


30 


3° 


30 


30 


30 


30 


7 


25 


42 


42 


42 


42 


40 


35 


30 


30 


30 


30 


30 


30 


30 


8 


25 


42 


42 


42 


42 


42 


38 


36 


32 


30 


29 


28 


26 


25 


9 


25 


42 


42 


42 


42 


42 


38 


34 


32 


30 


29 


28 


26 


25 


10 


25 


42 


42 


42 


42 


42 


38 


33 


32 


30 


29 


28 


26 


25 


11 


25 


42 


42 


42 


42 


42 


38 


33 


31 


29 


28 


27 


25 


24 


12 


25 


42 


42 


42 


42 


42 


37 


32 


30 


28 


27 


26 


24 


23 


13 




42 


42 


42 


42 


42 


37 


32 


30 


27 


25 


24 


22 


20 


14 




42 


42 


42 


42 


40 


35 


30 


28 


26 


25 


23 


22 


20 


14! 




42 


42 


42 


42 


36 


33 


30 


28 


25 











In the tables of angles the areas are given, and the weights in 
pounds per foot of section can be obtained by multiplying the 
areas by 3.4. 

The moments of inertia and location of the centers of gravity 
of the angles are given in the tables following. 

The radii of gyration are readily found from 



-fi 



I = inertia; 
A = area in square inches. 



All data are based upon dimensions in inches and weights in 



pounds 



14 



GRAPHICS AND STRUCTURAL DESIGN 



DIMENSIONS AND AREAS OF ANGLES 



Uneven 
legs. 


Even 
legs. 


1 

8 


A 


1 

4 


5 
T6 


3 
8 


A 


1 
2 


9 

T6 


I 


11 
T6 


3 

4 


\\ 


7 

8 


\\ 


I 




8 X8 
6 X6 














775 

5-75 

5-00 
4-75 
4-50 
4.25 
4.00 
3-75 
3.5o 
3.25 
3.00 
2.75 
2.50 
2.25 
2.00 


8.68 
6.43 
5-59 
5-31 
5-03 
4-75 
4-47 
4.18 
3.90 
362 
3-34 
3- 06 
2.78 


9.61 
7. 11 
6.17 
5-86 
555 
5-23 
4.92 
4.61 
4.30 
3.98 
3.67 
336 


10 ? 


11. 4 12.3 
8.44 9-OS 
7.31 7-8'/ 
6.94 7-4'/ 
6.56 7. of 
6.19 6.6= 
581 6.25 
5-44 5.84 
5-o6 5.4; 
4.69 50; 
4-31 4.6s 


13.2 

9-74 
8.42 
7-99 

7-55 
7. 11 
6.67 








14. 1 

10.4 
8.97 
8.50 
8.03 


15.0 












4.36 


5.06 
4.40 
4.18 
3.97 
3.7s 
353 
331 
309 
2.87 
2.65 
2.43 
2.22 
2.00 
1.78 
1.56 
1.30 


7 
6 
6 
6 
5 
5 
5 
4 
4 
4 
3 


78 
75 
41 
06 
72 
37 
03 
68 
34 
00 
65 


*7 X3? 










9-5o 
9.00 
8.50 


6 X4 


*S X5 










3.61 
3.42 
3.23 
3-05 

2.86 
2.67 
2.48 
2.30 
2. 11 
1.92 
1-73 
1-55 
1.36 
1. 17 
0.99 


6 X3s 










5 X4 












5 X3h 










2.56 
2.40 
2.25 
2.09 
1.93 
1.78 
1.62 
1-47 
1. 31 

I. IS 

1. 00 
0.84 




5 X3 


4 X4 








*4 X3j 








4 X3 


3iX3i 








3iX3 
3s X 2\ 








3 X3 

*2jX2f 

2IX2J 

*2^X2| 
2 X2 

ifXif 






1.44 
1. 31 

1. 19 
I.06 
0.94 
0.8l 
O.69 


3 X2^ 
3 X2 






2^X2 


0.36 


0.81 
0.72 
0.62 
o.S3 



Special sections. 



MOMENTS OF INERTIA AND CENTERS OF GRAVITY OF 

ANGLES 

Even Legs 



Thick- 


1 


3 


1 


5 


3 


7 


I 




ness. 


4 


8 


2 


8 


4 


8 




Legs. 


I 


g 


I 


g 


7. 


g 


I 


g 


I 


g 


I 


g 


/ g 




8 X8 










48.6 


2.2 


59-4 


2.2 


69.7 


2.3 


79-6 


2.3 


89.O 2. 


4 


6 X6 






15-4 


1.6 


19.9 


1-7 


24.2 


1.7 


28.2 


1.8 


31 


9 


1.8 


35-5 I. 


9 


*5 X5 






8.7 


1.4 


11. 3 


1.4 


13.6 


1.5 
















4 X4 






4-4 


1.1 


5.6 


1.2 


6.7 


1.2 


7-7 


1.3 


8 


6 


1.3 






3*X3§ 






2-9 


1.0 


3.6 


I.I 


4-3 


I.I 


5-0 


1.2 


5 


5 


1.2 






3 X3 


1.2 


0.84 


1.8 


0.89 


2.2 


0.93 


2.6 


0.98 
















*2|X2f 


0.95 


0.78 


1-3 


0.82 


1-7 


O.87 




















2§X2| 


0.70 


0.72 


0.98 


0.76 


1.2 


O.81 




















2|X2i 


0.50 


0.65 


0.70 


0.70 
























2 X2 


o.35 


0.59 


0.48 


0.64 


0.59 


O.68 



















* Special sections. 



MATERIALS 



IS 



MOMENTS OF INERTIA AND CENTERS OF GRAVITY OF 

ANGLES 

Uneven Legs 



r\ 



a.. 

T 



Br- 



Thick- 




i 






1 




5 




3 


7 




ness. 


Axis. 


4 




r 


2 




8 




1 


8 


I 


Legs. 




I 


g 


I 


g 


I 


g 


I 


g 


I 


g 


J 


g 


I 


g 


*7 X3l 


[1 










4-4 


0.78 


5-3 


0.82 


6.1 


0.87 


6.8 


0.91 


7-5 


0.96 










25.4 


2.5 


30.9 


2.6 


36.0 


2.6 


40.8 


2.7 


45-4 


2.7 


6 X4 


II 






4.9 


0.94 


6.3 0.99 


7-5 


1.0 


8.7 


1.1 


9-8 


I.I 


10.7 


1.2 






13. S 


1.94 


17.4 


1.99 


21. 1 


2.03 


24.5 


2.1 


27.7 


2.1 


30.8 


2.2 


6 X3* 


M 






33 


0.79 


4-3 


0.83 


5.1 


0.88 


5-8 


0.93 


6.6 


O.97 


7-2 


1.0 






12.9 


2.0 


16.6 


2.1 


20.1 


2.1 


23.3 


2.2 


26.4 


2.2 


29.2 


2.3 


*5 X4 


u 






4-7 


1.0 


6.0 


1.1 


7-1 


1.1 


















8.1 


1.5 


10.5 


1.6 


12.6 


1.6 














5 X3* 


n 






3.2 


0.86 


4-1 


0.91 


4-8 


0.95 


'5.6 


1.0 


"t.2 


1.0 










7-8 


1.6 


10. 


1-7 


12.0 


1.7 


13-9 


1.7 


15.7 


1.8 






5 X3 


(i 






2.0 


0.70 


2.6 


o.75 


3-1 


0.80 


35 


0.84 


3-9 


0.88 










7-4 


1-7 


9-5 


1.8 


II. 4 


1.8 


13.2 


1.8 


14.8 


1-9 






*4 X3^ 


■5 






3.o 


0.96 


3-8 


1. 00 


4-5 


1. 00 


















4.2 


1.2 


53 


1.2 


6.4 


1.3 














4 X3 


{I 






1.9 


0.78 


2.4 


O.83 


2.9 


0.87 


3-3 


0.92 


3-7 


0.96 










4.0 


1.3 


5-1 


1-3 


6.0 


1.4 


6.9 


1-4 


7-8 


1.5 






3*X3 


U 






1.9 


0.83 


2.3 


0.88 


2.8 


0.92 


3-2 


O.96 


35 


1.0 










2.7 


1.1 


35 


1.1 


4.1 


1.2 


4-9 


1.2 


5-2 


1-3 






3iX2| 


S 


O.78 


0.61 


1.1 


0.66 


1.4 


0.70 


1.6 


o.75 


1.8 


0.79 










1.8 


1.1 


2.6 


1.2 


32 


1.2 


3-9 


1.3 


4-4 


1-3 










3 X2\ 


s 


0.74 


0.66 


1.0 


0.71 


1.3 


0.75 


1.5 


0.79 














1.2 


0.91 


1.7 


0.96 


.2.1 


I.O 


2.5 


1.0 














3 X2 


h 


0.39 


0.49 


o.S4 


0.54 


0.67 


O.58 


















1.1 


0.99 


1.5 


1.0 


1.9 


I.I 


















2^X2 


{i 


o.37 


O.S4 


0.51 


0.58 


0.64 


O.63 


















0.65 


0.79 


0.91 


0.83 


1.1 


0.88 



















Special sections. 



The properties of angles of intermediate thickness can be 
interpolated from the tables with sufficient accuracy for most 
purposes. 



i6 



GRAPHICS AND STRUCTURAL DESIGN 



PROPERTIES OF STANDARD CHANNELS 


























Radius of 




Weight 
per 
foot. 


Flange, 
b. 


iWeb, 
t. 


Gauge, 
m. 


Tan- 
gent, 
T. 


Max. 
bolt or 
rivet. 


Moment 
of in- 
ertia, 

axis 1— 1. 


Moment 
of in- 
ertia, 

axis 2-2. 


Dis- 
tance 
base 
toe. 


Area. 


gyration. 




Axis 


Axis 




















of g. 




1-1. 


2-2. 


r 


55-00 


3.82 


0.82 


2.50 


12.25 


>| r 


43o 


12.2 


0.82 


16.18 


5.16 


0.87 




50.00 


372 


0.72 


2.50 


12.25 




403 


11. 2 


0.80 


14.71 


523 


0.87 


1 


45-O0 


362 


0.62 


2.00 


12.25 


- \ « 


375 


10.3 


0.79 


13.24 


532 


0.88 


40.00 


352 


0.52 


2.00 


12.25 


348 


9-4 


0.78 


11.76 


5-44 


0.89 




3500 


343 


0.43 


2.00 


12.25 




320 


8.5 


o.79 


10.29 


5-57 


0.91 


I 


33-00 


3.40 


0.40 


2.00 


12.25 


J 


312 


8.2 


0.79 


9.90 


562 


0.91 


r 


40.00 


3-42 


0.76 


2.00 


10.00 


1 r 


197 


6.6 


0.72 


11.76 


4.09 


0-75 


J 


35 00 


330 


0.64 


2.00 


10.00 




179 


5-9 


0.69 


10.29 


4.17 


0.76 


30.00 


3.17 


0.51 


1-75 


10.00 


y i < 


162 


5-2 


0.68 


8.82 


4.28 


0.77 


I 


25.00 


305 


0.39 


1-75 


10.00 


j i 


144 


4-5 


0.68 


7-35 


4-43 


0.78 


20.50 


2.94 


0.28 


1.75 


10.00 


128 


3-9 


0.70 


6.03 


4.61 


0.81 


r 


35.00 


3.18 


0.82 


1-75 


8.25 


1 r 


116 


4.7 • 


0.69 


10.29 


335 


0.67 




30.00 


3.04 


0.68 


1-75 


8.25 




103 


4.o 


0.65 


8.82 


3.42 


0.67 


H 


25.00 


2.89 


0.63 


1-75 


8.25 


Y i i 


9i 


3-4 


0.62 


7-35 


3.52 


0.68 




20.00 


2.74 


0.38 


1-50 


8.25 


1 


79 


2.9 


0.61 


5-88 


3-66 


0.70 


I 


1500 


2.60 


0.24 


1.50 


8.2s 


J I 


67 


2.3 


0.64 


4.46 


3.87 


0.72 


r 


25.00 


2.81 


0.61 


1.50 


7.25 


1 r 


71 


3-0 


0.62 


7-35 


3.10 


0.64 


20.00 


2.65 


o.45 


150 


7-25 


}•{ 


61 


2.5 


0.58 


5.88 


3.21 


0.65 


1 


1500 


2.49 


0.29 


1.38 


7-25 


51 


2.0 


0.59 


4.41 


3.40 


0.66 


I 


1325 


2.43 


0.23 


1.38 


725 


47 


1.8 


0.61 


3.89 


349 


0.67 


f 


21.25 


2.62 


0.58 


1.50 


6.25 


1 1 


48 


2.3 


o.59 


6.25 


2.76 


0.60 


18.75 


2.53 


0.49 


1.50 


6.25 


44 


2.0 


o.57 


5.5i 


2.82 


0.60 


8-{ 


16.25 


2.44 


0.40 


1.50 


6.25 


r * 1 


40 


1.8 


0.56 


4.78 


2.89 


0.61 




1375 


2.35 


0.31 


1.38 


6.25 




36 


1.6 


0.56 


4.04 


2.98 


0.62 


L 


11.25 


2.26 


0.22 


1.38 


6.25 


J I 


32 


1.3 


0.58 


3-35 


3.10 


0.63 


f 


19.75 


2.51 


0.63 


150 


5-50 


1 f 


33 


1.8 


0.58 


5.81 


2.39 


0.56 




1725 


2.41 


o.53 


150 


5.50 


M 


30 


1.6 


o.55 


5.07 


2.44 


0.56 


H 


14.75 


2.30 


0.42 


1.50 


5-50 


27 


1.4 


o.53 


4-34 


2.50 


0.57 




12.25 


2.20 


0.32 


125 


5.50 


J 1 


24 


1.2 


o.53 


360 


2.59 


0.57 


L 


9-75 


2.09 


0.21 


1.25 


5.50 


21 


0.98 


0.55 


2.85 


2.72 


o.59 


r 

1 


I5.50 


2.28 


0.56 


125 


4.50 


] r 


19.5 


1.3 


0.55 


456 


2.07 


o.53 


13.00 


2.16 


0.44 


125 


4.50 


L , i 


173 


1.1 


0.52 


3.82 


2.13 


0.53 


10.50 


2.04 


0.32 


125 


4.50 


1 f 1 


15. 1 


0.88 


0.50 


3.09 


2.21 


o.53 


L 


8.00 


1.92 


0.20 


1. 13 


4.5o 


J I 


13.0 


0.70 


0.52 


2.38 


2.34 


0.54 


r 


11.50 


2.04 


0.48 


1. 13 


3-75 


1 r 


10.4 


0.82 


0.51 


3.38 


1.75 


0.49 


H 


9.00 


1.89 


0.33 


1. 13 


3-75 


r * i 


8.9 


0.64 


0.48 


2.65 


1.83 


o.49 


I 


6.50 


1.75 


0.19 


1. 13 


3-75 


J I 


7-4 


0.48 


0.49 


1.95 


1.95 


0.50 


r 


7.25 


1.73 


0.33 


1. 00 


2.75 


1 , f 


4.6 


0.44 


0.46 


2.13 


1.46 


0.46 


4 i 


6.25 


1.65 


0.25 


1. 00 


2.75 


r * i 


4-2 


0.38 


0.46 


1.84 


1. 51 


0.45 


I 


5.25 


1.58 


0.18 


1. 00 


2.75 


J I 


3.8 


0.32 


0.46 


1.55 


1.56 


o.4S 


3J 


6.00 


1.60 


0.36 


0.88 


1.75 


1 r 


2.1 


0.31 


0.46 


1.76 


1.08 


0.42 


5.00 


1.50 


0.26 


0.88 


1.75 


r * 1 


1.8 


0.25 


0.44 


i.47 


1. 12 


0.41 


L 


4.00 


1. 41 


0.17 


0.88 


1-75 


J I 


1.6 


0.20 


o.44 


1. 19 


1. 17 


0.41 



Note. — This table is taken from the handbook of the Cambria Steel Co. 



MATERIALS 



17 



PROPERTIES OF STANDARD I BEAMS 




Si 


u 

II 


11 

§0 


8 . 

5 v 

•*" is- 


*o . 


"0 . . 

SI" 

6 §.a 




*o a • 
W ° 7 


C m. OT 


to. 2 | 


to 

"C-S 






Q* 


3**"" 


<% 


e~° 


$* 


J~S 


«E« 




o-*« 2 










d 




A 


t 


b 


1 


5 


r 


r 


r' 




n 


T 


Ins. 


Lbs. 


Sq. ins. 


In. 


Ins. 


Ins.- 1 


Ins.3 


Ins. 


Ins. 4 


In. 


Ins. 


Ins. 


Ins. 


3 


5.5o 


1.63 


0.17 


2.33 


2.5 


1.7 


1.23 


0.46 


0.53") 








3 


6.50 


1. 91 


0.26 


2.42 


27 


1.8 


1. 19 


o.53 


0.52^ 


I 


ift 


ili 


3 


750 


2.21 


0.36 


2.52 


2.9 


1-9 


I-I5 


0.60 


0.52J 








4 


7-So 


2.21 


0.19 


2.66 


6.0 


30 


1.64 


0.77 


0.59^1 








4 


8.50 


2.50 


0.26 


2.73 


6.4 


32 


1. 59 


0.85 


0.S8 I 


i 


I* 


A\ 


4 


9- SO 


2-79 


0.34 


2.81 


6.7 


3-4 


1.54 


0.93 


0.58 f 


4 


10.50 


3.09 


0.41 


2.88 


7-1 


3-6 


1. 52 


1. 01 


o.57j 








5 


9-75 


2.87 


0.21 


300 


12. 1 


4-8 


2.05 


1.23 


0.65I 








5 


12.25 


360 


0.36 


3. is 


13.6 


5-4 


1.94 


1-45 


0.63^ 


* 


1! 


3f 


5 


14-75 


434 


0.50 


329 


IS- 1 


6.1 


1.87 


1.70 


0.63J 








6 


12.25 


3.61 


0.23 


333 


21.8 


73 


2.46 


1.85 


0.72I 








6 


14.75 


434 


0.35 


345 


24.0 


8.0 


2.35 


2.09 


0.69^ 


I 


2 


4ft 


6 


1725 


5.07 


0.47 


357 


26.2 


8.7 


2.27 


2.36 


0.68 J 








7 


1500 


4-42 


0.25 


3-66 


36.2 


10.4 


2.86 


2.67 


0.78I 








7 


17-50 


5.15 


0.35 


3.76 


39-2 


11. 2 


2.76 


2-94 


0.76!- 


t 


2i 


5i 


7 


20.00 


5.88 


0.46 


3.87 


42.2 


12. 1 


2.68 


324 


o.74 J 








8 


18.00 


533 


0.27 


4.00 


56.9 


14.2 


3.27 


3.78 


o.84l 








8 


20.25 


5.96 


o.35 


4.08 


60.2 


150 


3.18 


4-04 


0.82 1 


i 


2\ 


6ft 


8 


22.75 


6.69 


0.44 


4.17 


64.1 


16.0 


3.10 


4.36 


0.81 f 


8 


25.25 


7-43 


0.53 


4.26 


68.0 


17.0 


3.03 


4.71 


0.80 J 








9 


21.00 


6.31 


0.29 


4-33 


84.9 


18.9 


3.67 


5.16 


0.90*1 








9 


25.00 


7-35 


0.41 


4-45 


9i-9 


20.4 


354 


5.65 


0.88! 


I 


2i 


7ft 


9 


30.00 


8.82 


0.57 


4.61 


101.9 


22.6 


3.4o 


6.42 


0.85 r 


9 


35.00 


10.29 


o.73 


4-77 


in. 8 


24.8 


3.30 


7-31 


0.84 j 








10 


25.00 


7-37 


0.31 


4.66 


122. 1 


24.4 


4.07 


6.89 


0.97") 
0.93 1 








10 


30.00 


8.82 


0.45 


4.80 


134.2 


26.8 


3.90 


7.65 


1 


2f 


7ii 


10 


35.00 


10.29 


0.60 


4-95 


146.4 


293 


3-77 


8.52 


0.91 f 
0.90 j 


* 


10 


40.00 


11.76 


0.75 


5-10 


158.7 


31.7 


3.67 


9-50 








12 


31.50 


9.26 


o.35 


S.00 


215.8 


36.0 


4.83 


9- So 


1. oil 
0.99 > 








12 


35.00 


10.29 


0.44 


5.09 


228.3 


38.0 


4.71 


10.07 


A 


2f 


s>i 


12 


40.00 


11.76 


0.56 


5. 21 


245-9 


41.0 


4-57 


10.95 


0.96 J 








IS 


42.00 


12.48 


0.41 


5. 50 


441.8 


58. 9 


5.95 


14.62 


1.08"! 
1.07 | 








15 


45.oo 


13.24 


0.46 


5-55 


455-8 


60.8 


5. 87 


1509 








IS 


50.00 


14.71 


0.56 


5.65 


483.4 


64. 5 


573 


16.04 


1.047 


! 


3 


12J 


IS 


55.00 


16.18 


0.66 


5-75 


5H. 


68.1 


562 


17.06 


1.03 1 
1. 01 J 








IS 


60.00 


17.65 


o.75 


5.84 


538.6 


71.8 


5.52 


18.17 








18 


55.0 


15.93 


0.46 


6.00 


795-6 


88.4 


7-07 


21.19 


1. IS"*] 


0.875 






18 


60.0 


17.65 


0.56 


6.10 


841.8 


93- S 


6.91 


22.38 


1. 13 1 


1 






18 


65.0 


19.12 


0.64 


6.18 


881.5 


97-9 


6.79 


23.47 


I.llf 


1 


31 


isft 


18 


70.0 


20.59 


0.72 


6.26 


921.2 


102.4 


6.69 


24.62 


I.09J 


1 






20 


65.0 


19.08 


0.50 


6.25 


1169.5 


117. 


7.83 


27.86 


1.21*1 








20 


70.0 


20.59 


0.58 


6.33 


1219.8 


122.0 


7.70 


29.04 


1.19^ 


1 


3* 


i6il 


20 


75.o 


22.06 


0.65 


6.40 


1268.8 


126.9 


7.58 


30.25 


I.I? J 








24 


80.0 


23.32 


0.50 


7.00 


2087.2 


1739 


9.46 


42.86 


1.36*1 

1. 331 








24 


85.0 


25.00 


o.57 


7.07 


2167.8 


180.7 


9.31 


44-35 








24 


90.0 


26.47 


0.63 


7.13 


2238.4 


186.5 


9.20 


45.70 


1. 31 *> 


1 


4 


20IS 


24 


95.o 


27.94 


0.69 


7.19 


2309.0 


192.4 


9.09 


47.io 


1.30 | 








24 


100. 


29.41 


0.75 


7.25 


2379.6 


198.3 


8.99 


48.55 


1. 28 J 









Note. — This table is taken from the handbook of the Cambria Steel Co. 



i8 



GRAPHICS AND STRUCTURAL DESIGN 



Grey Mill Sections 

The Bethlehem Steel Company on their Grey Mill are able 
to roll sections with much wider flanges than are possible with 
the usual methods of manufacture. These sections make better 
columns and can be used upon longer spans without lateral brac- 
ing. The following tables give some of these sections with their 
properties. In H columns, under each section-number, only the 
first three weights and the maximum weight are given although a 
large number of intermediate weights are rolled. The beam and 
girder-beam sections are given almost completely. 



H COLUMNS 

,Y 

- «i 



'" — ir— • 



u 





. 


a 












AjusX^ 




Axis YY. 


6 


1 

u 


.2 





















2 














a 
_o 


p. 


0) 

03 


D 


b 


* 


T 


*o 


« 52 


"o eJ • 


*o 


CJ w 


c . 


tn 




1 

< 










11 


§.2 

<u 


lis 

Pi 60 


si 


§.3 

0) 


05 O W 

'■dgo 

rt 60 




Lbs. 


Sq. ins. 










1 


5 


r 


r 


S' 


r' 


f 


83-5 


24.5 


13! 


139 


0.43 


11. 1 


884.9 


128.7 


6.01 


294.5 


42.3 


3-47 


H 14 | 


91.0 


26.8 


135 


14.0 


0.47 


11. 1 


976.8 


140.8 


6.04 


325.4 


46.6 


3-49 


99-0 


29.1 


14 


14.0 


0.51 


11. 1 


1070.6 


1530 


6.07 


356.9 


51.0 


350 


I 


287.5 


84.5 


i6| 


14.9 


1. 41 


11. 1 


3836.1 


454-7 


6.74 


1226.7 


164.7 


3.81 


( 


64- 5 


19.0 


nf 


11. 9 


0.39 


9.2 


4990 


84.9 


5.13 


168.6 


28.3 


2.98 


H 12 | 


7i. 5 


21.0 


Hi 


12.0 


o.43 


9.2 


556.6 


937 


5.15 


188.2 


31.5 


3.00 


78.0 


22.9 


12 


12.0 


0.47 


9.2 


615.6 


102.6 


5.18 


208.1 


34.7 


301 


I 


161. 


47-3 


13* 


12.5 


0.94 


9.2 


1444.3 


214.0 


553 


477.0 


76.5 


3.18 


f 


40.0 


14.4 


9i 


10. 


0.36 


7-7 


263.5 


53-4 


4.28 


89.1 


17.9 


2.49 


H io \ 


54-0 


15.9 


10 


10. 


0.39 


7-7 


296.8 


59-4 


4.32 


100.4 


20.1 


2.51 


59-5 


17.6 


ioi 


10. 


0.43 


7-7 


331.9 


65.6 


4-35 


112. 2 


22.3 


2.53 


I 


1235 


36.3 


11* 


10.5 


0.86 


7-7 


790.4 


137-5 


4.67 


2593 


49-5 


2.67 


( 


31-5 


9.2 


7i 


8.0 


0.31 


6.1 


105.7 


26.9 


3.40 


35.8 


8.9 


1.98 


H %\ 


34-5 


10.2 


8 


8.0 


0.31 


6.1 


121. 5 


30.4 


3.46 


41. 1 


10.3 


2.01 


39-0 


11. 5 


8| 


8.0 


o.35 


6.1 


139.5 


34-3 


3.48 


47.2 


II. 7 


2.03 


1 


90 5 


26.6 


9h 


8.5 


0.78 


6.1 


385.3 


81. 1 


3.80 


125. 1 


29.6 


2.17 



MATERIALS 



19 



GIRDER BEAMS 
















— D — 




— n 










0? 


1. 


Is 

11 




3 . 


Neutral axis perpen- 
dicular to web at 
center. 


Neutral axis 
coincident 
with center 
line of web. 


+j 

4) • 

>8 

i-.cS 






.a 


I s 


"BE 


1 i 

« 


8 
«>.2 


0) 

M 6 


hi 

1 -s 


a 

to O 

3+3 


Is 






D 






( 


6 


1 


r 


I 
e 


r 


r' 




M 


T 


30 


200.0 


58.71 


0.750 


15.00 


9150.6 


12.48 


610.0 


630.2 


3.28 




II.OO 


25.2 


30 


180.0 


53.00 


0.690 


1300 


8194.5 


12.43 


546.3 


433.3 


2.86 




9-00 


25.2 


28 


180.0 


52.86 


0.690 


1435 


7264.7 


11.72 


518.9 


5333 


3.18 




10.25 


23.4 


28 


165.0 


43. 47 


0.660 


12.50 


6562 . 7 


11.64 


468.8 


3719 


2.77 




8.50 


234 


26 


160.0 


46.91 


0.630 


13 60 


5620.8 


10.95 


432.4 


435.7 


305 




9-50 


21.6 


26 


150.0 


43-94 


0.630 


12.00 


5153.9 


10.83 


396.5 


3146 


2.68 




8.00 


21.6 


24 


140.0 


41.16 


0.600 


13.00 


4201.4 


10.10 


35o.i 


346.9 


2.90 




9.00 


20.0 


24 


120.0 


35.38 


0.530 


12.00 


3607.3 


10.10 


300.6 


249.4 


2.66 




8.00 


20.3 


20 


140.0 


41.19 


0.640 


12.50 


2934.7 


8.44 


2935 


348.9 


2.91 




8.50 


15.7 


20 


112. 


32.81 


0.550 


12.00 


2342.1 


8.45 


234.2 


2393 


2.70 




8.00 


16.4 


18 


92.0 


27.12 


0.480 


11.50 


1591-4 


7.66 


176.8 


182.6 


2.59 




7-50 


14.8 


15 


I40.0 


41.27 


0.800 


11.75 


1592.7 


6.21 


212.4 


331.0 


2.83 




7-75 


10. 1 


IS 


IO4.O 


30.50 


0.600 


11.25 


1220 . 1 


6.32 


162.7 


213.0 


2.64 




725 


11. 1 


15 


73.0 


21.49 


0.430 


10.50 


883.4 


6.41 


117. 8 


123.2 


2.39 




6.50 


12. 1 


12 


70.0 


20.58 


0.460 


10.00 


538.8 


5.12 


89.8 


II4-7 


2.36 




6.00 


90 


12 


55-0 


16.18 


0.370 


975 


432.0 


5-17 


72.0 


81. 1 


2.24 




6.00 


95 


10 


44.0 


12.95 


0.310 


9.00 


244-2 


4-34 


48.8 


573 


2.10 


J 


5-50 


7-8 


9 


38.0 


11.22 


0.300 


8.50 


170.9 


3.90 


38.0 


44.1 


1.98 


* 


5.25 


9-9 


8 


32.5 


954 


0.290 


8.00 


114. 4 


3.46 


28.6 


32.9 


1.86 


¥ 


5.00 


6.0 



BEAMS 



30 


120.0 


353 


0.54 


10.5 


5239.0 


12.2 


349-0 


165.0 


2.2 


I 


6.50 


26.4 


28 


105.0 


30 


9 


0.50 


10. 


4014.0 


II. 4 


287.0 


131. 


2.1 


I 


6.00 


24-7 


26 


90.0 


26 


s 


0.46 


9-5 


2977-0 


10.6 


229.0 


IOI.O 


1-9 


I 


5-50 


230 


24 


84.0 


24 


8 


0.46 


9-3 


2382.0 


9-8 


198.0 


91.0 


1.9 


5 


5-25 


21.0 


24 


73-0 


21 


S 


o.39 


9-0 


2091 . 


99 


174.0 


74.0 


1-9 


i 


5-25 


21.3 


20 


82.0 


24 


2 


o.57 


8.9 


1560.0 


8.0 


156.0 


80.0 


1.8 


i 


5.00 


17. 1 


20 


72.0 


21 


4 


o.43 


8.7 


1466.0 


8.3 


146.0 


76.0 


1.9 


I 


5.00 


17. 1 


20 


69.0 


20 


3 


0.52 


8.1 


1269.0 


7-9 


127.0 


51.0 


1.6 


* 


4.50 


17- 5 


20 


64.0 


18 


9 


0.45 


8.1 


1222.0 


8.0 


122.0 


50.0 


1.6 


i 


4-50 


17- 5 


20 


59-0 


17 


4 


0.38 


8.0 


1172.0 


8.2 


117. 


48.0 


i-7 


A 


4-50 


17-5 


18 


59-0 


17 


4 


0.50 


7-7 


883.0 


7-1 


98.0 


39.0 


1.5 


* 


4.25 


15-7 


18 


54.0 


15 


9 


0.41 


7-6 


842.0 


7-3 


94-0 


38.0 


1.5 


1 


4-25 


15-7 


18 


52.0 


15 


2 


0.38 


7.6 


825.0 


7-4 


92.0 


37.0 


1.6 


? 


4.25 


15-7 


18 


48.5 


14 


2 


0.32 


7-5 


798.0 


75 


89.0 


36.0 


1.6 


4-25 


157 


15 


71.0 


20 


9 


0.52 


7-5 


796.0 


6.2 


106.0 


61.0 


i-7 


4 


4-25 


11. 7 


15 


64.0 


18 


8 


0.60 


7.2 


665.0 


6.0 


89.0 


42.0 


1-5 


I 


4.00 


12.3 


15 


54-0 


IS 


9 


0.41 


7.0 


610.0 


6.2 


81.0 


38.0 


1.6 


* 


4.00 


12.3 


15 


46.0 


13 


S 


0.44 


6.8 


485.0 


6.0 


65.0 


25.0 


1.4 


I 


3-75 


12.9 


15 


41.0 


12 





0-34 


6.7 


457-0 


6.2 


61.0 


24.0 


1.4 


4 


3-75 


12.9 


15 


38.0 


II 


3 


0.29 


6.7 


443.o 


6.3 


59.o 


23.0 


1-4 


i 


375 


12.9 


12 


36.0 


10 


6 


0.31 


6.3 


269.0 


5.0 


45.o 


21.0 


1.4 


i 


3-50 


9-9 


12 


32.0 


9 


4 


o.33 


6.2 


228.0 


4.9 


38.0 


l6.0 


1-3 


1 


3-50 


10.2 


12 


28.5 


8 


4 


0.25 


6.1 


216.0 


5.1 


36.0 


15.0 


1-3 




3-50 


10.2 


10 


28.5 


8 


3 


0.39 


6.0 


134-0 


4.0 


27.0 


12.0 


1.2 


i 


3-25 


8-4 


10 


23.5 


6 


9 


0.25 


5-9 


123.0 


4.2 


25.0 


II. 


1-3 


i 


3-25 


8.4 


9 


24.0 


7 





0.36 


5-6 


92.0 


3.6 


20.0 


9-0 


i.l 


4 


3-00 


7.5 


9 


20.0 


6 





0.25 


5-4 


85.0 


3.8 


19.0 


8.0 


1.2 


i 


300 


7-5 


8 


195 


5 


8 


0.32 


5-3 


61.0 


3.2 


15.0 


7-0 


I.I 


i 


2.75 


6.6 


8 


17.5 


5 


2 


0.25 


52 


57-0 


3.3 


14.0 


6.0 


I.I 


i 


2.75 


6.6 



CHAPTER II 
GRAPHICS 

Statics. — Statics treats of forces at rest and therefore in 
equilibrium, hence the resultant force in any direction and the 
moments of the forces about any point must be zero. For 
coplanar forces the condition of equilibrium is expressed in the 
following equations: 

2 horizontal forces = o, 
2 vertical forces = o, 
2 moment of forces about any point = o. 

Graphic Statics. — Graphic statics relates to the solution of 
statical problems by geometrical constructions. 

Force. — Force is an action upon a body tending to change 
its state of rest or motion, A force is completely known when 
its magnitude, direction, line of action and point of application 
are known. 

Magnitude. — Forces may be measured by any unit of weight. 
It is most convenient in this work to use the pound. The magni- 
tude of a force can be represented graphically by the length of 
a line, the length being drawn to a previously chosen scale, for 
instance, if the scale is i inch to iooo pounds then a line J-inch 
long represents a force of 500 pounds. 

Direction. — An arrow placed on the line is used to indicate 
the direction of the force. 

Line of Action. — The line of action is along the line repre- 
senting the force and it is along this line that the force tends to 
produce motion. 

Point of Application. — The point of application is the place 
assumed as a point where the force acts upon the body. 



GRAPHICS 21 

Coplanar and Noncoplanar Forces. — Coplanar forces have 
their lines of action in a common plane, while the lines of action 
of noncoplanar forces do not He in the same plane. Where not 
otherwise stated coplanar forces are to be understood. 

Concurrent and Nonconcurrent Forces. — Concurrent forces 
have their lines of action intersect in a point, while the lines of 
action of nonconcurrent forces fail to meet in a common point. 

Equilibrium. — A number of forces are in equilibrium when 
they produce no tendency to motion in the body upon which 
they act. 

Couple. — Two equal and parallel but opposite forces con- 
stitute a couple. The arm of the couple is the perpendicular 
distance between their two lines of action. 

Moment of a Couple. — The moment of a couple is the product 
of either force by the arm of the couple. 

Resultant. — The resultant of a system of forces is the single 
force or simplest system that would produce the same effect as 
the other forces and could, therefore, replace them. 

Equilibrant. — The equilibrant is equal in magnitude to the 
resultant but opposite to it in direction. It is, therefore, the 
single force or the simplest system that will exactly neutralize 
the given system of forces. 

Let the force Fi acting on the point d be represented in direc- 
tion and magnitude by the line ad, 
and similarly the force F 2 , also act- 
ing on d, be represented by dc, then 
completing the parallelogram and 
drawing the diagonal db gives R, the 
magnitude and line of action of the F 

resultant. Its direction, to produce 

the same effect as the two forces Fi and F 2 , must be as indicated 
by the arrow. 

Force Triangle. — The triangle dbc would have given the 
resultant force as well as the parallelogram. Suppose the two 
known forces are laid off so that the arrows run in the same 





22 GRAPHICS AND STRUCTURAL DESIGN 

direction, that is either clockwise or counterclockwise; then the 

b line closing the triangle with its ar- 
row following around in the same 
direction as the other two arrows, 
here counterclockwise, will be equal 
to the resultant but opposite in di- 
rection; hence, E is the force which 
would hold the two forces Fi and F 2 in equilibrium and is 
their equilibrant. 

Force Polygon. — If more than three forces meet at a point and 
are in equilibrium a force polygon may be constructed by finding 
the resultant of two of the forces, combining this resultant with 
a third force to find a second resultant, and so on until all the 
forces have been considered and a final resultant determined. 





Fig. 3. 



Ri is the resultant of F 3 and F 4 ; R 2 is the resultant of Ri and 
F 5 , and F 2 is the equilibrant of the forces F h F b , F± and F z . 

It is evident that in the construction the drawing of the 
resultants Ri and R 2 might have been dispensed with, it only 
being necessary to lay the forces off so that the arrows would 
follow in the same direction. It follows then that any number 
of concurrent coplanar forces will be in equilibrium if their force 
polygon closes and that any side of the force polygon represents 
the equilibrant of all the other forces. 

Equilibrium of Coplanar, Nonconcurrent Forces. — In the 
case of coplanar nonconcurrent forces the closing of the force 
polygon is not alone sufficient proof of equilibrium. This will be 
seen from the following simple example. The three forces F h F 2 
and F 3 are equal and act in the same plane making an angle of 



GRAPHICS 



23 



120 degrees with each other. It is evident that, being an 
equilateral triangle, the force polygon closes. For the system 
to be in equilibrium, however, the equilibrant of F 2 and F 3 
should coincide with E. It is evident that the equilibrant of the 
system would be a clockwise moment Fi X a. Hence, when 
a system of nonconcurrent forces is in equilibrium the force 





Fig. 5. 



Fig. 6. 



polygon must close and the sum of the moments of the forces 
must be zero. It follows that three nonconcurrent forces can- 
not be in equilibrium unless the forces are parallel and that the 
resultant of the group of nonconcurrent forces may be either 
a single force or a couple. 

Components. — In a system of forces having a resultant each 
force is a component of the resultant. Hence, a force may have 
any number of components. 

Composition of Forces. — Composition of forces consists in 
finding for a system of forces an equivalent system having a 
smaller number of forces. The simplest case of composition of 
forces is finding a single force replacing a system of forces. 

Resolution of Forces. — Resolution of forces consists in finding 
for a system of forces an equivalent system having a greater 
number of forces. An illustration of this is where a given force 
is resolved into two or more forces or components. 



Equilibrium Polygon 

A system of forces is shown in Fig. 7. Fig. 8 is the force 
polygon. Fi and F 2 intersect and the fine of action of their 
resultant must pass through their point of intersection and be 



24 



GRAPHICS AND STRUCTURAL DESIGN 



parallel to a, the resultant of F x and F 2 in the force polygon. 
Similarly, the resultant b oi[a and F 3 must pass through the point 
of intersection of a and F 3 in Fig. 7, and be parallel to b in the 
force polygon, Fig. 8. In this way we finally reach R, the result- 
ant of F i} with b, the resultant of all preceding forces F h F 2 and 





Fig. 8. 



F 3 . In Fig. 7, R is located by the fact that it must pass through 
the intersection of F 4 and b. It must also be equal and parallel 
to R in Fig. 8. 

Figure 10 is an equilibrium or funicular polygon. 

In the above system the - force polygon, Fig. 9, closes but 
F 5 does not coincide with the resultant R of the other forces 




Fig. 9. 




Fig. 10. 



Fi, F 2 , Fz and F 4 , in the equilibrium polygon, hence, the sys- 
tem lacks equilibrium due to the clockwise moment F 5 X d. 
The system is in equilibrium for translation. The resultant 
of the system is F 5 X d. The equilibrant of the system is 
-F b Xd. 



GRAPHICS 



25 



The method just given applies when the several forces can be 
conveniently made to intersect. It should be noticed that an 
infinite number of equilibrium polygons are possible for a given 
system of forces. 

When the forces do not intersect conveniently the equilibrium 
polygon is drawn as follows: 

Take any point 0, called a pole, outside of the forces F h 
F 2 and F3 and connect the extremities of these forces with the 
lines called rays, a, b, c and d. The force Fi will be held in equi- 
librium by the two rays a and b, acting as indicated by the 
arrows inside the triangle formed by the force F\ and the rays a 
and b. In the same way b and c hold F 2 , and c and d hold F 3 in 
equilibrium, also R, the resultant of F\, F 2 and F 3 , is held in 



y^-\ 






Fig. 11. 




Fig. 12. 



equilibrium by the rays a and d. Now if two forces represented 
by the rays a and b hold F± in equilibrium, then in the equilib- 
rium polygon these three forces must intersect in a point. Take 
any point on Fi in Fig. 11, and through it draw the lines called 
strings parallel respectively to b and a in Fig. 12. F 2 is held in 
equilibrium by the rays b and c; therefore, in Fig. 11, produce b 
until it intersects the force F 2 and through this point of inter- 
section draw the string c parallel to the ray c, Fig. 12. Through 
the point of intersection of string c and force F 3 in Fig. 1 1 draw 
the string d parallel to ray d in Fig. 12 and produce it until it 
cuts string a. Now, since the resultant R, Fig. 12, is held in 
equilibrium by the rays a and d, these three forces R, a and d 
must intersect in a point in Fig. 11. The intersection of a and 
d locates R in Fig. 11, its magnitude and direction being given 
by Fig. 12. 



26 



GRAPHICS AND STRUCTURAL DESIGN 



Figure 12 is the force polygon. Fig. n is the equilibrium 
polygon. The difference between these two polygons should be 
carefully noted. The force polygon gives the direction and 
magnitude of the forces while the equilibrium polygon gives the 
lines of action and the direction but not the magnitude. 

Graphic Moments 

Let Fi, F 2 , F 3 and F 4 be four forces constituting a system whose 
bending moment is desired about a point p. Draw the force 





Fig. 13. 



Fig. 14. 



polygon (Fig. 14), take a pole 0, draw the several rays and draw 
the pole distance Os perpendicular to the resultant R; this dis- 
tance is H. In Fig. 13 draw the equilibrium polygon making the 
strings parallel to the rays in Fig. 14. Through the intersection 
of strings a and e draw the resultant R. Through the point p 
draw y parallel to R and limited by the string e and the string a 
produced. The bending moment of R about the point p is 
RX h. The triangles 123 and 4 O 5 are similar, hence 

R:H::y:h or RXh = H X y. 

That is, the bending moment of any system of coplanar forces 
about any point in the plane is the product of the pole distance 
H, and the line y, drawn through the point p, parallel to the 
resultant of the forces R and limited by the two strings e and 
a which intersect on the resultant. 



GRAPHICS 



2 7 



Moment of Parallel Forces. — The equilibrium polygon may 
be used to determine the moment of any or all of a system of 
parallel forces. 

The bending moment on the beam at the section y-y due to 
the forces R l and F x at the left of that section is equal to 1-2, 



u— - 9- 



F _h 



~^--i 



Fig. 15. 



-Am 



-r 6 r 

-I 












J< H— 

Fig. 16. 



the intercept in the equilibrium polygon multiplied by the pole 
distance H. Expressed algebraically the bending at y-y equals 

M =(R 1 Xg) - (Fi X h). (1) 

Now triangles 135 and 680 are similar, hence 1-5 :g::6-8:H. (2) 

Also triangles 245 and 670 are similar, so that 2-5 : h: : 6-7 : H. (3) 

In equation (2), 6-8 = R h therefore Ri X g = 1-5 X H. 

In equation (3), 6-7 = F h therefore F x X h = 2-5 X H, and 

M = (1-5 X H) - (2-5 X H) = 1-2 X H. 

The bending moment at any point on a beam due to a system 
of parallel forces is equal to the ordinate of the equilibrium 
polygon, cut off by a line drawn through the given point and 
parallel to the resultant of all the forces multiplied by the pole 
distance. By bending moment on any beam section is meant 
the algebraic sum of the moments to the left of that section. 
The intercept in the equilibrium polygon is a distance and should 
be measured to the same scale as that to which the beam is laid 
off. The pole distance is a force and should be measured by the 
scale used in the force polygon. . 



28 



GRAPHICS AND STRUCTURAL DESIGN 



Uses of Force and Equilibrium Polygons 

Reactions of a Beam. — A supported beam of span L carries 
a load F a distance a from the left support. 

Lay off F in Fig. 18, take the pole and draw the rays a and b. 
Through any point 2 on line of action of load F in Fig. 17 draw 
the two strings parallel to the rays a and b. Through the points 
1 and 3 where the rays a and b respectively cut the lines of the 





us a — -> 

1 


F 

1 




n 




t 


» 


1 


\ 


Ri 






Rt- 



Fig. 17. 



-I- 
1. 



Fig. 18. 



>o 



reactions Ri and i? 2 draw the string c. Now in Fig. 18 draw the 
ray c through the pole and parallel to the string c in Fig. 17. 
This ray c cuts the force F into two parts representing the two 
reactions, Ri being held in equilibrium by the rays a and c and R 2 
by the rays c and b. 

Problem. — In Fig. 19 find the reactions when 
F\ = 10 tons (20,000 lbs.). 
F 2 = 2 tons (4000 lbs.). 
Answer. — Ri = 9300 lbs. R 2 = 14,700 lbs. 

Problem. — Draw force and equilibrium polygons for the 
beam in Fig. 20, finding reactions Ri and R 2 and the maximum 
bending moment. Check the answers by algebraic calcula- 
tions. 

Problem. — Draw force and equilibrium polygons for the beam 
in Fig. 21, find the reactions R x and R 2 and the maximum bend- 
ing moment. Check the answers by algebraic calculations. 

In treating uniform loads graphically it is necessary to divide 
the load into small sections and the load corresponding to each 
section is considered at its center. See Figs. 22 and 23. 



GRAPHICS 



29 



±± 



Fig. 19. 



1 



L _4L-> *_ _ 5 1 _s, ^SL — 6- -^ 



•■ *8 



ft!* 



ffe * -sfjl 



Fig. 20. 



I 



ke 20- sJ 

Fig. 21. 



a: 





Force Polygon 

Fig. 23. 



H--8^ 



F 



Fig. 24. 



Fig. 25. 



J S B k 



-L. 



18V 



— 14- 



\ ! 8* n ^>k3 



%VJ 

^ 






■f 

11 



A 

Fig. 27. 




Fig. 26. 



30 GRAPHICS AND STRUCTURAL DESIGN 

Problem. — The beam shown in Fig. 24 carries a uniform 
load of 1000 lbs. per foot of length; draw the shear and bending- 
moment diagrams, using the force and equilibrium polygons. 

Determination of the Center of Gravity. — Divide Fig. 25 
into convenient regular sections, in this case three rectangles. 
Assume forces acting through the centers of gravity of these 
rectangles representing their areas. Then, by means of force 
and equihbrium polygons (Figs. 26 and 27), locate the resultant 
or equilibrant of these three forces. The location of the vertical 
equilibrant is shown. 

Similar treatment of these areas taken as acting horizontally 
will give the horizontal equilibrant and the center of gravity of 
the figure will lie at the intersection of these two equilibrants. 

Deflection oe Beams 

For beams carrying irregular loadings the graphical deter- 
mination of- deflections is convenient. Books on mechanics of 
materials deduce the general equation of the elastic curve of 

beams as -j% = — • M is the bending moment in inch pounds. 
ax 1 EI 

If the equihbrium polygon representing the bending moment 
upon the beam is known the curve of its deflection may be drawn 
in the following way : 

The given bending-moment diagram is divided into sections 
and the areas of these sections are represented by the lines A, B, 
C, etc., in Figs. 28 (a) and (d). In the funicular polygon, the pole 
O is taken a distance to represent EI, E being the modulus of 
elasticity of the material of the beam, expressed in pounds per 
square inch, and / the moment of inertia of the section, in inches 4 , 
and referred to that axis of the beam about which the deflec- 
tion occurs. Fig. 28 (c) is an equilibrium polygon drawn in the 
usual way for the two Figs. 28 (a) and 28 (d). The intercepts y 
in the diagram, when properly scaled, give the deflection at that 
point on the beam. To determine the scale if 1 inch measured 
horizontally in Fig. 28 (a) represents 5 inches of span, 1 inch in 



GRAPHICS 



31 



Fig. 28 (b) equals P pounds, H the pole distance in Fig. 28 (b) 
measured in inches, H 1 the pole distance of Fig. 28 (d) also 
measured in inches, further in Fig. 28 (d) 1 inch equals k square 
inches of the bending moment area of Fig. 28 (a). Then 1 inch 



of intercept in Fig. 28 (c) represents 



k-P .s*.H -H 1 
E.I 



inches. 






r 




i 


\. (b) 


1 


Force ^v 
Polygon ^v 


LL 




Vector 
Polygon 



Fig. 28. 



The correctness of this method may be shown by taking a 
section of the moment curve and dividing it into strips having 
a width dx. The area of one of these sections is M- dx. Com- 
paring the similar triangles Org and the infinitesimal triangle, 

M -dx d 2 y M d 2 y _,. . , , 

I his is the general equation 



we have 



d 2 y M 

Tx or Y~ 



E-I dx E-I dx 2 
of the elastic curve, as well as the equation of the line in 
Fig. 28 (c). 



32 



GRAPHICS AND STRUCTURAL DESIGN 



Continuous Beams. — Another use of the curves of deflec- 
tion is the solution of restrained or continuous beams with 
irregular loads and spans. 

The following is a modification of the method credited to 
Dr. Geo. Wilson, " Proceedings of the Royal Society," Vol. 62, 
Nov., 1897. 



U--i L -d 



Force and Vector 




Fig. 29. 

The method consists of plotting the bending moment and< 
deflection curves due to the external loads assumed as carried by 
the outer supports. Similar curves are then drawn for the same 
beam excepting that now the actual loads are removed and as- 
sumed reactions are applied at the intermediate supports, the 
beam being held as before at the outside reactions. If the beam 
is level the deflections caused by the first loading must be equal 
and be opposite to those produced by the second loading. Equat- 



GRAPHICS 33 

ing these values of the deflections at the several supports gives 
sufficient data for the determination of the reactions and when 
these are known, a revised moment diagram for the actual beam 
is readily made. The solution is simplified by using the same 
pole distance for all the force or vector polygons, and using the 
same scale for the forces in all diagrams. An example will show 
the method in detail. 

A continuous beam, Fig. 29 (a), carries three loads on four sup- 
ports. The loads are, AB = 1000 lbs. ; BC = 800 lbs. ; and CD = 
1400 lbs. The lengths of the spans beginning with that at A 
are 12 ft., 10 ft. and 8 ft. The loads are placed as shown. 

The curve of deflection Fig. 29 (e) is first found for a beam sup- 
ported at the ends and carrying the three loads, AB, BC and CD. 
The force polygon Fig. 29 (c) is drawn, from which the bending 
moment diagram readily follows in Fig. 29 (b). This polygon 
Fig. 29 (b) is then divided into a number of small sections of 
uniform width, similar to the one cross-hatched. Since they 
have the same widths their areas may be represented by their 
middle ordinates, shown heavy in the cross-hatched section and 
marked 5. These middle ordinates may now be laid off in the 
vertical line of the vector polygon Fig. 29 (d). From this vector 
polygon the deflection curve Fig. 29 (e) is obtained in a manner 
similar to that used for the bending-moment diagram Fig. 29 (b). 

Under the support GF the deflection in Fig. 29 (e) scales 51, 
while under FE it scales 40. 

Now removing the loads AB, BC and CD, assume a unit load 
GF applied below the beam and acting along the line of the 
reaction. As before construct a force polygon Fig. 29 (f), the 
equilibrium polygon of bending-moment diagram Fig. 29 (g) 
and finally the deflection curve Fig. 29 (i). In this polygon the 
deflections for the unit load at the lines of the reaction scale 27 
and 19. In the same way place a unit load at EF and draw the 
polygons /, k and m. The deflections over the reactions are 19 
and 18. The reactions will be some multiples of the unit loads 
applied. These unit loads may be any convenient one as 1, 



34 GRAPHICS AND STRUCTURAL DESIGN 

ioo, 500 or 1000 lbs., depending upon the beam and its loading. 
If we call the reaction GF, p times the unit load and FE, q 
times the unit load, then the deflections due to GF will be 27 X p 
and 19 X p, respectively, while those due to FE will be 19 X q 
and 18 X q. Since the deflections at the points of support due 
to the reactions equal the deflections due to the loads we have 
in this problem, under GF 

(27 X p) + (19 X q) = 51 
and under FE 

(19 X p) + (18 Xq) = 40. 

Solving, these equations give p = 1.26 and q = 0.89. The unit 
load at the reactions having been taken at 1000 lbs., the 
reactions become GF = 1.26 X 1000 = 1260 lbs., while FE = 
0.89 X 1000 = 890 lbs. 

The reaction AG may be found by taking moments about the 
right support ED. 

^ G< _ (1000X24) + (800X1 5) + (1400X4)- (890X8) -(1260X18) 

30 
= 390 lbs. 

Since the sum of the reactions equals the sum of the loads the 
reaction ED equals (1000 + 800 + 1400) — (390 + 1260 + 890) 
= 660 lbs. 

It should be noted that the pole distances were all equal and 
the same scale of forces was used in all force polygons. When all 
the forces and reactions are known the bending moments may be 
found either graphically by constructing the force and equilib- 
rium polygons as usual, or the moment at any section may be 
found by computing the moment of the forces acting on one side 
of the section. 



CHAPTER III 
STRESSES IN STRUCTURES 

The application of force and equilibrium polygons to the 
determination of stresses in structures is fairly simple, espe- 
cially if care is taken in properly marking the structure acted 
on by the external forces and the lines of the diagram as drawn. 

In the simple truss, Fig. 31, having located the forces at the 
several points, place a letter in each triangle of the truss and a 
letter between each pair of external forces. A member in the 



a) 


Upper Chord 
\ "Web Members/ ^^^ 


L 


Lower Chord 
Span >i 

Fig. 30. 




Fig. 32. 




Fig. 33 



Fig. 35. 



structure will be designated by the letters on each side of it, 
while the stress in the member will be designated in the stress 
diagram by the same letters at its extremities. The truss being 
symmetrical and symmetrically loaded the reactions will be equal 
and, since the sum of the vertical forces must be zero, each re- 
action must equal one-half the total load, hence the reaction is 

A „ BC + CD + DD' + D'C + C'B' 

A3 = • 



35 



36 GRAPHICS AND STRUCTURAL DESIGN 

To assist the explanation, numbers have been placed at some 
of the points but this is not usually required. Considering the 
forces acting at apex 1, AB and BC are known and the resultant 
of these is held in equilibrium by forces whose lines of action are 
CE and EA . The values of these two forces can, therefore, be 
found by drawing the force polygon of the four forces acting at 
point 1. This has been done in Fig. 32. It is important to 
place the directions on these forces and then indicate these 
directions as acting to or from their point of application. This 
has been done at apex 1. As soon as an arrow is placed at 1 
on CE the arrow can be placed on CE at apex 2, since it must be 
opposite in direction to the first arrow. Similarly, the arrow on 
EA can be placed at apex 3. Now considering the forces acting 
at apex 2, EC and CD are known and, therefore, the other two, 
DF and FE, being known in direction, can be found in magnitude 
as in Fig. 33. 

Again taking the forces at apex 3, we now know EF and EA 
from which FG and GA can be found. This has been done in 
Fig. 34. In this manner, force polygons could be drawn for all 
the apices. 

Generally, instead of drawing these force polygons separately 
as just done, they are superimposed upon each other making a 
diagram, Fig. 35. Where the truss is symmetrical about a ver- 
tical axis through the center of its span and the corresponding 
loads on the two sides are equal, it is only necessary to make 
the portion of the diagram shown in full lines. The other half 
is shown dotted and by rotation about AE could be super- 
imposed upon the full-line diagram. In this case, drawing this 
dotted diagram is unnecessary. 

Tension or Compression in the Member 

Where the forces acting upon the structure are indicated as 
shown, a piece having the arrows < — > shows that piece to be in 
compression, while — ><— indicates tension. 



STRESSES IN STRUCTURES 



37 



Note. — In analyzing stresses in this way the members at the apices 
are assumed as pin connected and the external forces of the pieces are indi- 
cated by the arrows as acting on the pin. That is, the forces acting on the 
pin and not the forces of the pin on the pieces are indicated by the arrows. 

Frequently trusses carry loads attached to the lower chords. 
The solution graphically follows the procedure just described. 

If the lower apex loads BC, CD, etc., are equal the reactions 
AB and AB' will be equal and each will be one-half the sum of 
the loads on the lower chord. Taking the forces at point i, we 
can lay off AB, knowing both its direction and magnitude, then 
laying off BF and FA parallel to the respective truss members 



i 


4 


A 

e 
i/ 


! 
K / 

/l 


3 

\k' 
■l'\ 


X 

J'\ 


\ 


F*S 


B 


3 

\ c i 


5 

t D N 


7 

' E 1 


I Ef 


D' \ 


' C' ' 


*' 




Fig. 36. 



Fig. 37. 



the stresses BF and FA become known in magnitude. Now 
going to point 2, since AF and AG will be parallel lines passing 
through a common point A they will coincide and their intercept 
on FG will be a point or its value is zero. Hence the stress in 
FG due to loads carried on the lower chord is zero. 

The points F and G will be coincident on Fig. 37. Now going 
to point 3, GF, FB and BC are known, hence CH and EG are 
readily drawn and their magnitudes determined. 

The method of determining the direction of the forces is the 
same as previously explained and, as before, the members in the 
upper chord are shown to be in compression, those in the lower 
chord in tension. 

The vertical members, excepting LL\ will be in compression 
while the diagonal members are in tension. 

Moving Loads Carried under Trusses. — Not infrequently 
trolley or hoist runway tracks are carried by the lower chords 
of trusses. In this event the track is preferably fastened to \he 



38 



GRAPHICS AND STRUCTURAL DESIGN 



lower chord at the apices of the triangles, and the load may come 
upon any apex. It will generally be easy under these circum- 
stances to make a diagram for the load under each apex in one- 
half the lower chord, and by comparing the diagrams, which are 
preferably made to the same scale, the maximum stress in any 
member due to any of the several positions of the load is readily 
found. A diagram made for one position of the load will illus- 
trate the method. 





F-G 

Fig. 40. Fig. 42. 

The load is assumed at apex 3. The reactions AB and AB f must 
first be found. Draw the force polygon, Fig. 41, and the equi- 
librium polygon, Fig. 39. When the ray 3, in Fig. 41, is drawn 
parallel to the closing side of the equilibrium polygon, Fig. 39, 
the load BB r is divided into the two reactions. 

When it is required to find the reactions for all positions of the 
load, the work can be abridged by laying off the load BB f at the 
left reaction and taking any point Q in the other reaction and 
drawing BQ and B'Q, then the intercept a-a! in the triangle 
under the load will be the left reaction for that position of the 
load. If BQ and B'R are drawn parallel to the lower chord a-a! 
will be the left reaction and a-b the right reaction. 

Note. — The student should prove the truth of this last diagram, using 
both algebraic and graphical demonstrations. 



STRESSES IN STRUCTURES 



39 



Wind Load on Trusses 

The wind load will be assumed normal to the left side of the 
roof and both ends of the truss will be fixed. 

The reactions AB and AB' will be parallel to the wind loads 
BC, CD, DE and EF. The sum of the reactions will equal the 
sum of the loads. The magnitude of the reactions can be found 
by the force and equilibrium polygons, Figs. 44 and 45. In 
Fig. 44, BB' is laid off parallel to the wind loads and equal to 
their sum. The resultant normal wind pressure on the roof 
Fig. 43. 




Fig. 46. 

equals the sum of the wind loads BC, CD, DE, EF and FB f and 
will act coincident with the force DE. Take any point p in the 
line of this resultant and through it draw the strings parallel to 
their respective rays 1 and 2 in Fig. 44. Then in Fig. 45 draw 
the closing line of the equilibrium polygon and in Fig. 44 draw 
ray 3 parallel to string 3 in Fig. 45. BB' will then be divided 
into two parts corresponding to the two reactions. Having found 
the two reactions the stress diagram follows naturally as in 



40 GRAPHICS AND STRUCTURAL DESIGN 

Fig. 46. No difficulty is experienced until apex 3 is reached. 
Here there are three forces known completely and three known 
in direction only. It is, therefore, impossible to determine the 
magnitude of the unknown forces unless more conditions are 
assumed. One solution of the problem is as follows. It can be 
shown that the forces acting in FL, LM and MA will be the 
same whether the truss is left as shown by the full lines JK and 
K.L, or if these are removed and replaced by the member shown 
dotted, JL. 

Taking moments about point 5, cutting the truss as shown by 
the line and equating internal and external moments we have 

21 external moments = (force FL) X a.* 
As the forces LM and MA pass through point 5, their moments 
about 5 are zero. Hence, the force in FL depends only upon the 
external moments and the lever arm a, and is consequently inde- 
pendent of the truss members to the left of the cut. In a simi- 
lar way stresses in LM and MA can be shown to be independent 
of truss members to the left of the cut. Having found the tem- 
porary forces acting at 3, when JK is removed complete the dia- 
gram for the forces at 5 and 6, then remove the dotted member JL 
and replace the full members LK and KJ. Now, taking the forces 
acting at 6, EF and FL are known completely; therefore, the 
magnitudes of EK and KL can be found. From here taking the 
forces at point 7 and then returning to point 3 all the forces act- 
ing in the left half of the truss become known. The diagram can 
be closed by finally taking the forces acting at the right reaction. 

The stresses due to the wind in the members U'G 1 to L'M in 
the right half of the truss will be seen to be zero. It is evident 
that for the wind acting on the right side of the truss, the mem- 
bers in the right half will be stressed the same as similar members 
in the left half when the wind acts as shown in Fig. 46 ; hence, 
redrawing the diagram is unnecessary. 

In trusses of long span it may be necessary to have one end 
free to allow for expansion; this may be done by allowing the 
* Note. — Consult Method of Moments. Chapter IV. 



STRESSES IN STRUCTURES 



41 



truss to rest upon a plate secured to the wall. The plate fastened 
to the lower chord of the truss has slotted holes which permit 
the truss to slide upon the wall plate by overcoming the friction 
between the plates. The friction may be greatly reduced by 
using rollers under the truss. The following example illustrates 
the method of determining the reactions and stresses due to wind 
loading when the truss is free at one end. The plates are placed 
at the right-hand end of the truss. The coefficient of friction is 
assumed as one- third; hence the tangent a = J or the inclination 





Fig. 48. 



Fig. 47 



Fig. 49. 

is one in three. The line of action of the right reaction is, 
therefore, known. The first part of the problem requires the 
determination of the magnitude of the right reaction and of both 
the magnitude and the direction of the left reaction. It is 
known, however, that the left reaction must pass through point p 
of the truss. First draw the force polygon, Fig. 48. Through 
B f draw a line parallel to the right reaction. Take any pole 
and draw the rays 1, 2, 3 and 4. In Fig. 47, draw the cor- 
responding strings. 

Since the left reaction AB, the force BC and rays 1, 2 and 5 
are in equilibrium they must intersect in a common point in the 
equilibrium polygon; this must be point p, the only point com- 
mon to BC and the left reaction AB. Hence, through p draw a 
string parallel to ray 1. From Fig. 48 the force BC is held in 
equilibrium by the strings 1 and 2 ; hence, in Fig. 47 they must 



42 



GRAPHICS AND STRUCTURAL DESIGN 



have a common point. Through p draw also the string 2 parallel 
to ray 2. Force CD and rays 2 and 3 are in equilibrium and 
must pass through a common point in the equilibrium polygon, 
Fig. 47. From Fig. 48 rays 4 and 5 hold reaction AB' in equilib- 
rium and rays 1 and 5 hold reaction AB in equilibrium; hence, 
string 5 must pass through the point of intersection of string 4 
with reaction AB ' and it must also pass through the intersection 
of string 1 with reaction AB which is in point p. It is noticed 
that string 1 becomes merely a point in Fig. 47. This work 
would have been simplified by using a resultant wind pressure 
equal to BB', acting in the same line as CD in Fig. 47. 




Fig. 50 




Fig. 52. 

The stress diagram is shown in Fig. 49, its method of construc- 
tion differing in no way from those previously described. It 
should be noted that the diagram is for the entire truss, the 
loading being unsymmetrical, and that the diagram checks if the 
force B'E' when drawn through the points B' and E' in Fig. 49 
is parallel to the line B'E' of the truss, Fig. 47. 

The diagram should be constructed for the wind on the free 
side and such construction is shown in Figs. 50 to 52. 

When rollers are used at the free end the reaction at that end 
is assumed vertical. 

It is shown, page 54, how the stresses may be obtained 
algebraically by the method of moments and how the moments 



STRESSES IN STRUCTURES 



43 



due to external loads may be found graphically. A combination 
of these two methods may be used thus: 

Figure 56 is the force polygon for the external loads, with pole 
and the rays drawn. From this the equilibrium polygon, 
Fig. 55, has been drawn. Now the bending moment at any 
section is the product of the intercept under that section in the 
equilibrium polygon multiplied by the pole distance H. Hence 
to determine the stress in any member DF cut the truss at Z-Z 
and replace the cut members by the forces that would have to 
act in them to produce equilibrium. Now, if possible, take 



Fig. 53. 



Fig. 54. 





Fig. 55. 

moments about a point common to all but one of these forces, in 
this case point 1. Let a be the lever arm of the force DF about 
point 1, then making the sum of the internal and external mo- 
ments zero we have 



from which 



(HXy) + {DFXa) = o, 

DF= - (ffxy) . 

a 



The Character of the Stress, whether tension or compression, 
can be determined as follows: 

Calling clockwise rotation positive and counterclockwise neg- 
ative find the character of the moment of the external forces 
H X y. This is found to be positive. Since the sum of the 
internal and external moments about a point is zero the moment 



44 



GRAPHICS AND STRUCTURAL DESIGN 




of the internal forces must be of opposite character to the mo- 
ment of the external forces hence DF X a is negative and DF 
must act in the opposite direction to that assumed. This makes 
DF in compression. 



STRESSES IN STRUCTURES 45 

Stresses in a Bent 

The determination of the dead-load and snow-load stresses in 
the truss of a bent presents no greater difficulties than that of a 
truss carried upon brick walls, as there are no stresses in the knee 
braces due to these loads. Where the wind load is carried by the 
columns and truss forming the bent the magnitude of the forces 
acting in the truss members and of the stresses acting in the 
column section vary greatly with the distance from the top of 
the column to the foot of the knee brace and with the manner 
of securing the column to the foundation. 

The wind upon the truss may be considered as either horizontal 
or normal but will here be assumed as normal. The columns will 
first be considered as hinged top and bottom and then as fixed 
at the base and hinged at the top. In this latter case it can be 
shown that the point of contraflexure lies between the base of 
the column and the foot of the knee brace and between one- 
half and five-eighths of this distance from the base of the 
column. It is generally assumed as one-half for convenience, 
and this is sufficiently accurate. To expedite the determi- 
nation of the stresses the forces acting upon the column have 
been transferred to the truss through the extra truss mem- 
bers added temporarily to the columns. It should be noted 
that vertical sections can be taken through all members of 
the truss without cutting these added pieces, hence the stresses 
in these members will be independent of the stresses in the 
added pieces and the diagram drawn with their assistance will 
give the correct stresses in the permanent members of the 
truss. The wind load has been assumed as 12 lbs. per sq. ft., 
normal to the roof surface. The following dimensions have also 
been used: span 36 ft. o ins., bay widths 16 ft., base of column 
to lower chord of truss 14 ft., lower chord of truss to foot of 
knee brace 5 ft. o ins. 

In Case I the columns are hinged both top and bottom, while 
in Case II they are fixed at the base and hinged at the top. The 



4 6 



GRAPHICS AND STRUCTURAL DESIGN 



lettering has been made identical in both cases and, the scales 
being the same, the influence of the method of securing the 
columns upon the stresses in corresponding members can be seen 
at a glance by comparing the stresses in the stress diagrams, 
Figs. 59 and 62. The following explanation refers to Case I, but 
applies also to Case II. It is first necessary to estimate the apex 
wind loads on the sides and roof. On the side of the building 



1280* 640 




1440 * 



Fig. 60. 



Fig. 61. 



CASE II 
COLUMN BASE FIXED, 
COLUMN TOP HINGED 




the wind is assumed as acting at the base of the column, the 
foot of the knee brace and the top of the column. The normal 
wind pressure on the roof has a resultant W N equal to the sum 
of apex loads EF, FG, GH and HB' } and acting centrally with 
this side of the roof as shown. 



STRESSES IN STRUCTURES 47 

* Similarly, the horizontal wind pressure We equals the total 
horizontal wind pressure between two bents and acts centrally 
upon the side of the building. Now these two wind forces acting 
on the building have a resultant acting through their point of 
intersection given in direction and magnitude in the force 
polygon, Fig. 58. Taking any point in this resultant Rw in 
Fig. 57, an equilibrium polygon can be drawn, since it is known 
that the resultants AB and AB' must pass through the column 
bases. In Case II these resultants pass through points in the 
columns midway between the base of the column and the foot 
of the knee brace. Having drawn the strings 1, 2, and 3 in 
Fig. 57, draw the rays 1 and 2 in Fig. 58, and their point of 
intersection Q will be the desired pole. The usual assumption 
is that the columns share the horizontal components of the wind 
forces equally. In Fig. 58 the horizontal component of the wind 
forces is BS, which has TV drawn perpendicular to it from its 
center. If, through the pole Q, ray 3 is drawn parallel to string 3 
in the equiHbrium polygon in Fig. 57 it will cut TV at the point 
of intersection of the resultants AB and AB' acting at the 
column bases. In Case II these resultants act on the columns 
midway between the bases and the knee braces. Having found 
the resultant wind forces acting on the columns the drawing of 
the stress diagram presents no unusual difficulties. To transfer 
the forces from the column bases to the truss members C7, //, 
JD, B'T ', I' J' and J'B' have been added. 

The stresses given by the diagram for the column AI, AI r , 
K ' J f and KJ will not be the correct stresses for the actual 
structure, but the stresses of all other truss members including 
the knee braces are the desired ones. In the drawings of the 
truss the members in compression have been drawn heavy. 

Maximum Bending due to Moving Loads. — It has been 
shown, page 26, how the equiHbrium polygon can be used to 

* Note. — In Case II the wind pressure on the side effecting the stresses 
in the truss is assumed as that from the point of contraflexure to the top 
of the column and the resultant acts midway between these two points. 



48 



GRAPHICS AND STRUCTURAL DESIGN 



determine the bending moments upon a beam. By a simple 
extension of the principles a diagram can be made showing the 
maximum bending that will occur along the beam as a system of 
moving loads passes over the span. 

The loads BC, CD, DE and EF in Fig. 63 are a constant dis- 
tance apart and roll across the girder ab. The several positions 
of the system of loads relative to the girder can either be shown 
by redrawing several additional positions of the loads or by 
redrawing the girder, moving the girder under the loads. The 



\g B 



c JdJ e If 

<h 0)0)0) 





Fig. 63. 



Fig. 64. 



latter method, being the easier, has been used and two additional 
positions of the girder c-d and e-f have been drawn. Fig. 64 is 
the force polygon for the four wheel-loads, a pole was chosen 
and the rays 1 to 5 inclusive were drawn. 

Draw the equilibrium polygon for the several loads, the strings 
being drawn parallel to their respective rays. Strings 1 and 2 will 
intersect on the force BC, strings 2 and 3 will intersect on force 
CD, etc. Now draw the closing line of the first equilibrium poly- 
gon by drawing gh through the points of intersection of the ex- 
tended reactions AB and AF of the girder. Then for the first 
position of the loading upon the girder the bending moment at any 
point on the girder will be the product of the intercept measured 
in feet to the scale of the span by the pole distance E measured 
by the scale of forces. In the same way move the girder to the 
second position c-d and locate the closing line of the second equi- 
librium polygon in the fine to. Proceed in this way until the loads 



STRESSES IN STRUCTURES 49 

have been rolled across the girder. It now remains to compare the 
bending moments at definite points on the girder for the several 
positions of the loads. Suppose it is desired to know the maxi- 
mum bending moments at p intervals across the girder, one would 
begin by comparing the intercepts in the several equilibrium 
polygons a horizontal distance p from the left end. Several of 
these ordinatesyi, kl and mn have been drawn. The comparison 
may be assisted by drawing a curve through the points of inter- 
section of these ordinates with the closing lines of their respective 
equilibrium polygons. 

This is shown by the curves RR and SS. 

The intercepts between the curves RR and 55 and the broken 
line of the equilibrium polygon will give the bending moments 
exactly only at the vertexes of the equilibrium polygons ; there- 
fore, when the maximum appears to be between vertexes, the 
ordinate should be checked by drawing an equilibrium polygon 
with the loads in a position to bring a vertex at the desired point. 

Diagram of Maximum Live -load Shears. — This will be ex- 
plained by a loading somewhat resembling the usual locomotive 
and train load. 

In making the diagram of maximum shears, Fig. 66, lay off 
the forces BC, CD, etc., and take a pole distance OB equal to 
the girder span. Draw the rays 1 to 10 and beginning at 
draw the strings 1 to 10 parallel to their respective rays and in ac- 
cordance with the usual method, i.e., CD in the force polygon is 
held in equilibrium by rays 2 and 3 ; hence, in the equilibrium 
polygon, strings 2 and 3 must intersect force CD in a common 
point. The side of the equilibrium polygon under the uniform 
load is a parabola and can be drawn tangent to the broken line. 
To understand the theory of the diagram draw the closing line OA , 
string 1 1 , of the equilibrium polygon ; this is also ray 1 1 of the force 
polygon and divides the line of the forces B-H into the reactions. 
B-A is the left reaction when the load BC is over the left support 
with the other loads as shown. Suppose the loads moved to the 
right a distance a or the same relative position of the loads 



5o 



GRAPHICS AND STRUCTURAL DESIGN 



and girder is more readily obtained by moving the span a dis- 
tance a to the left as designated by position 2. The closing 
line in this case is string 12 and the reaction is A/TV. As the 
loads travel to the right the shears under load BC can be obtained 
by measuring from point B the distance the load BC is to the right 
of the left reaction and at this point measuring the ordinate from 

OB to the broken line 2,3 to 10 of the equilibrium polygon. 

This of course must be measured by the same scale as that to which 
the forces are laid off. When the usual locomotive wheel loads are 



Fig. 65. 



B ( t/(t) D (t) E (t) 



1 1 

Span-Position* 2 . 



3 



s\ 




F-^ti— - 



Span-Positional- 

Span-Position^8 . . 

Fig. 66. 



preceded by a much lighter pilot wheel there may be a question 
as to whether the greatest shear at any section occurs when the 
pilot wheel, load BC, is at the section or the first driver, load 
CD, is there. In Fig. 66, measure the span marked position 3, 
produce string 2 until it cuts PQ, the line of the right reaction. 
From the point where string 2 cuts BH draw a heavy horizontal 
line to the left. To find the maximum shear a distance b from 
the left support draw the intercept c-d, a distance b from the line 
QP ; this will be the shear when the load CD is at the distance b 
from the left support. Compare c-d with the intercept e-f drawn 
between the broken fine of the equilibrium polygon and the 
horizontal line OB and at the distance b from the line of forces 



STRESSES IN STRUCTURES 



51 



HB. In this way the maximum shear may be found for any 
point. 

This method is applicable to the determination of the maximum 
stresses in truss members as well as in plate girders. 

The following illustrates the method applied to the determi- 
nation of the stresses in a Pratt truss. 




C D E F G 

Fig. 67. 



Example. — The Pratt truss, Fig. 67, spans 150 ft. It has 
6 panels and height of 30 ft. The loading is Cooper's E-60; see 
page 72, under Influence diagrams, for this loading. 

Find the stresses in BC, be and diagonal bC, locating the 
positions of loads for maximum stress and shear by influence 
diagrams and using equihbrium diagrams for determining the 
maximum bending moments and shears. 



CHAPTER IV 

ALGEBRAIC DETERMINATION OF STRESSES 

The conditions of equilibrium used in the determination of 
stresses are: 

i. Sum of the horizontal forces = o. 

2. Sum of the vertical forces = o. 

3. Sum of the moments of the forces about any point = o. 
The stresses in the members cut by the section a-a can be 

replaced' by forces F h F 2 and F 3 , equal to the respective stresses 




Fig. 68. 



Fig. 69. 



in these members. These forces F h F 2 and F s hold the portion 
of the truss to the left of the section, Fig. 68, in equilibrium and 
the stresses in the members may be determined by using con- 
ditions 1, 2 and 3. 

Referring to conditions 1 and 2, they should be understood to 
mean that the sum of the components along any line must 
= o, and the sum of the components along a line at right angles 
to the first line must = o. 

Frequently the work may be abridged by using the more gen- 
eral scheme. Forces and components acting to the right or up- 
wards are positive, those acting to the left or down are negative. 
Clockwise moments are positive; counterclockwise moments are 

52 



ALGEBRAIC DETERMINATION OF STRESSES 



53 



negative. To find the stresses in CE and EA in Fig. 70, cut 
the truss at a-a, then, by conditions 2 and 1, 

(AB - BC) + CE sin a = o, 

CE = - {AB - BC) cosec a, 

also — CE cos a + AE = or ^4£ = CE cos a; 

here AE is plus, acts to the right and produces tension. 



a 


E P \j[> 


/ G \ 


7\c' 


■ / 


<\ f 


A 


/ 




b' b 

Fig. 70. 

In a similar way to find the stress in DF or FE, cut the truss in 
section b-b. Considering the vertical components we have 

(AB -BC - CD) - DF- sin a + EF sin = o. 

Now considering the horizontal components 

+ EA - DF cos a- EF cos 



o. 



The solution of these equations will give 
the stresses DF and EF. The forces 
acting at a point may also be treated x 
by this method. Considering the forces b 
acting at point 1 in Fig. 72, and using 
condition 2 we have 

(AB - BC) + CE sin a = o 

CE = - ( AB ~ BC ) . 




Fig. 72. 



or 



sina 



The force CE acts down towards point 1. yl£ — CE cos a = o; 
AE, being plus, acts towards the right. The direction and char- 
acter of the forces can be checked by drawing the force polygon 
for the forces at the point. 



54 



GRAPHICS AND STRUCTURAL DESIGN 



Method or Moments 

The stresses in framed structures can be determined alge- 
braically by placing the sum of the external and internal moments 
equal to zero. 

To determine the stress in any member cut the truss by a 
section passing through the member whose stress is desired. 
Taking the portion of the truss to the left of the section replace 
the members by forces acting to the right of the section as shown 
in Fig. 74. Now select some point about which to take moments. 
If possible, take a point through which all the cut members pass, 
excepting the member whose stress is desired. In the example 




Fig. 73. 



Fig. 74. 



given, members FG and GA pass through point 3 so that moments 
will be taken about point 3. To determine the direction of the 
force acting in the member assume that it acts away from the cut 
section or to the right and that the piece is in tension. If the 
solution gives a plus value the assumption was correct; if the 
value is minus the piece is in compression. As before clockwise 
moments are positive, while counterclockwise moments are 
negative. 

Taking moments about point 3 in Figs. 73 and 74, we have 
Moments of external forces + moments of internal forces = o. 

[{AB - BC) b] - [CD X (b-d)] + DF X e = o, 

DF _ - [{AB - BC) b] + [CD x (ft - d)] 



ALGEBRAIC DETERMINATION OF STRESSES 



55 



The quantity (AB — BC) b will be much greater than CD (b — d) 
and DF will, therefore, be negative, and consequently the piece 
DF will be in compression instead of in tension as assumed. 

To determine the stress in FG take moments about point i, 
then 

(CDXd) -(FGxf)=o, 



FG= + 



CDXd 

f 



The value of FG, being plus, agrees with the assumption and the 
member is in tension. 



E 2 b c 4 -10 



Fig. 75. 

E Wtang 

-12 -12 -10 



Wsec. 6 




Fig. 77. 



In dealing with bridge trusses with parallel chords an abridg- 
ment of the method of resolution of forces is possible. 

In Fig. 76 it should be noted that all the stresses are multiples 
of one of the sides of the small triangle ETB, thus F-D = 
3 • T-B; E-F = 3 • E-T and E-D = 3 • E-B. E-D is the ver- 
tical shear in panel D, E-C that in panel C, and E-B that in 
panel B. E-B = load W. 



56 GRAPHICS AND STRUCTURAL DESIGN 

Expressed in terms of the apex loads W and the angle 0, E-T 
= W • sec 6 and T-B = W • tang 0, hence : 



Member. 


Stress. 


Member. 


Stress. 


E-F and F-G 

G-H and H-I 

I- J and J-K 

K-L 


3 W sec 
2 W sec 
i IF sec 
o W sec 


F-D 

E-G 

H-C 

I-E 

B-J 

K-E 


3 W tang 

6 IF tang 

8 IF tang 

io W tang 

ii IF tang 

12 IF tang 







It is seen that the coefficients of the diagonals are the coeffi- 
cients of the shears in their panels, thus the shear in panel D 
is 3 • W, hence 3 is the coefficient on E-F and F-G, being — for 
compression members and + for tension pieces. In the same 
way the shear in panel C is 2 • W, hence the coefficient is 2 for 
diagonals G-H and H-I. 

The next point to be noted is that the sum of the coefficients 
of the members cut by the sections a-a, b-b and c-c must in each 
case be zero. Section a-a cuts members E-F and F-D; the 
coefficients then are — 3+3=0. At section b-b the coefficients 
of D-F and F-G are each + 3. Now if the coefficient of E-G is x, 

then +3+3+* = °> 

hence x = — 6. 

In this way all the coefficients can be found and stresses com- 
puted by multiplying these coefficients by W • sec 6 for diagonal 
members or by W • tang 6 for horizontal chords. 

Live Loads. — To analyze the influence of live loads at the 
several apexes of the lower chord, construct Fig. 77. With the 
load at apex 3, the left reaction is f- P, the right reaction \ P. 
The coefficients of the web members to the left of the load, apex 3, 
are f while those to the right are — \. 

In the same manner from Fig. 78, which is the stress diagram 
for the load P at apex 5, it is seen that the coefficients of the 
web members to the left of apex 5 are y while to the right the 
coefficients are — f. 



ALGEBRAIC DETERMINATION OF STRESSES 



57 




58 GRAPHICS AND STRUCTURAL DESIGN 

It will be seen that for maximum chord stresses the truss should 
be fully loaded, while for minimum chord stresses there should be 
no live load on the truss. In this type of truss all loads on upper 
or lower chords produce tension in the lower chord members and 
compression in the upper chord members. 

The maximum web stress will be produced when the longer 
segment from that panel to a pier is fully loaded. The minimum 
web stresses will be created when the shorter segment from panel 
to pier is fully loaded. 

Pratt Truss 

The analysis of the stresses in a Pratt truss can be done by the 
method of coefficients in the same way as the Warren truss just 
described. If, as is sometimes done, parts of the dead loads are 
assumed as applied at the apexes of the upper chord the only 
members whose stresses will be affected are the vertical ones. 
To find the shear in these verticals cut the truss diagonally 
through the vertical member whose stress is desired and equate 
the shears. (See Fig. 83.) 

/-/ = (C-D) - (C-B) - (B-A) - (D-E). 

Had there been no loads upon the upper chords, as in Figs. 
81 and 82, then 

/-/ = (C-D) - (C-B) - (B-A). 

The maximum chord stresses are equal to the sum of the dead- 
and live-load chord stresses, while the minimum chord stresses 
are those due to the dead load only. The maximum and mini- 
mum web stresses are found by adding algebraically the corre- 
sponding live- and dead-load stresses. 

Since in this truss the diagonals can take only tension the 
necessity for counter diagonals to care for stresses due to un- 
symmetrical live loading should be noted. The stress in the 
counter is the same as the compression would have been in the 
piece for which it acts. 



ALGEBRAIC DETERMINATION OF STRESSES 



59 





Fig. 82. 




Fig. 83. 



6000 ~ 12000 „ 12000 _ 12000 _ Wtangfl 



Wseofl 




Fig. 84. 



60 GRAPHICS AND STRUCTURAL DESIGN 

Warren Truss 

Problem. — Find the dead-load stresses by the method of 
coefficients in the following deck Warren truss, Fig. 84. Span 
100 ft. Load at apexes on the upper chord, 12,000 lbs. 

The secant of 45 degrees is 1.414. Tangent of 45 degrees is 1. 

Shear in panel C = 3.5 X W — 42,000 lbs. 

Shear in panel D = 42,000 — 12,000 = 30,000 lbs. 

Shear in panel E = 42,000 — 12,000 — 12,000 = 18,000 lbs. 

Shear in panel F = 42,000 — 12,000 — 12,000 — 12,000 = 6000 lbs. 

Stress in HG = — 3I W seed = — 3.5 X 12,000 X 1.414 = — 59,400 lbs. 

Stress in HI = + 2\ W seed = + 2.5 X 12,000 X 1.414 = + 42,400 lbs. 

Stress in JK = — i| W seed = — 1.5 X 12,000 X 1.414 = — 25,500 lbs. 

Stress in KL = + \ Wsecd = + 0.5 X 12,000 X 1.414 = + 8,500 lbs. 

Stress in HA = +3.5 WtsmgO = + 3.5 X 12,000 X 1 = + 42,000 lbs. 

Stress in KA = -+- 7.5 PFtang0 = + 7.5 X 12,000 X 1 = + 90,000 lbs. 

Stress in DI and EJ = — 6 W tang 6 = — 6 X 12,000 X 1 = — 72,000 lbs. 

Stress in FL = — 8 W tang = — 8 X 12,000 X 1 = — 96,000 lbs. 

Stress in CG = o 

Stress in GB = — 6000 lbs. 



CHAPTER V 
INFLUENCE DIAGRAMS 

When a system of concentrated loads moves across a girder or 
a trussed bridge the maximum moment or shear at a section, or 
stress in a given member, will be produced by a certain position 
of the system of loads relative to that section or member. 

Influence lines are used to determine this position of the loads 
producing maximum moments, shears or stresses. 

Influence Diagrams. — An influence diagram shows the varia- 
tion of the effect at any particular point, or in any particular 
member, of a system of loads moving over the structure. In- 
fluence diagrams are commonly drawn for a load of unity. The 
moments, shears, or stresses for any system of loads can be com- 
puted from the intercepts in this diagram by multiplying them 
by the given loads. 

Influence diagrams to find the position of loading to give maxi- 
mum moment at a given point in a beam or girder or at a given 
joint on the loaded chord of a truss. 

In Fig. 85, 

2Pi is the resultant of moving loads to the left of point 3. 
2P 2 is the resultant of moving loads to the right of point 3. 

To construct the influence diagram, Fig. 86, for the bending 
moment at 3, compute the bending at 3 due to a unit load at 
this point 

ft. 'xis-fl and M = (A^. 

If p is laid off on the left reaction and (5 — p) upon the right 
reaction and their extremities are joined to the ends of the hori- 

61 



62 



GRAPHICS AND STRUCTURAL DESIGN 



zontal line ea by ba and/e then the vertical intercept cd represents 
(S-p)p 
S 

eab and dac, - = — —2- 

p s 

triangles dec and aef. 

G - P) P . 



for, let this intercept cd be x then by similar triangles 

: = ^ ^ and similarly for the 

o 



S-p 



or x = 



Fig. 85. 




I (s-i>) 



Fig. 86. 



To find the moment at a given point 3, due to a system of moving 
loads, multiply the intercept under each load by that load and 
take the sum of these products. From Fig. 86 

M = 2P1V1 + 2P 2 y2. 

Now move the system of loads a small distance dx to the right, 
allowing no load, however, to pass on or off the span or across 
the given point 3. Then for the new position 

M + dM = 2Pi(yi + dyj + 2P 2 (y 2 - dy 2 ). 

Subtracting the preceding equation from this we have 

dM = XPxdyi - 2P 2 dy 2 . 



INFLUENCE DIAGRAMS 63 

Now examining for maximum bending by placing this equation 
= o, we have 

2P 1 dy 1 = 2P 2 dy 2 . (1) 

By similar triangles 





d *- S Z p and dy / = 

dx dx 


.P 
" S 


from which 


dyi S - p 
dy 2 p 






Substituting 


this value in equation 


(1) 




hence 
from which 


ZPi (S - p) = 
2Pi5 - 2Pip 


= 2P2P, 





or 



SPiS = 2 (Pi + P^ p 



s f 1 _ s(f, + f 2 ) 



p s 

This may be expressed by stating that the maximum bending 
will occur at a section when the average load to the left of the 
section equals the average load on the entire span. 

The influence diagram to find the position of the loading to give 
a maximum moment at a given joint on the unloaded chord of a 
truss, having either parallel or inclined chords, may be found in a 
similar way. 

Figures 87 and && are drawn for joint 4 on upper chord. 

If the system of loads is moved a distance dx to the left the 
rate of change in the bending moment is 

^~ = -ZP 1 d yi - 2P 2 dy 2 + 2P 3 dy z • (1) 

rr u . dM 

To be a maximum —r- = o. 
dx 



6 4 



GRAPHICS AND STRUCTURAL DESIGN 



But by the geometry of the construction 

dyi _ S — p . dyz _ p , dj2 _ fS — pc 

dx S dx S dx cS 



Fig. 87. 



K— /-H 




Fig. 88. 

These values substituted in equation (1) give the position for 
maximum moment at joint 4. 



S 



SP 2 ^+ 2Pi 
c 



SP is sum of all loads Pi, P 2 and P 3 . 

Maximum Shear 

Reactions and Shears. — By definition, shear is the algebraic 

sum of the vertical forces to the left of the section. When the 

unit load has moved a distance x to the left of the right support, 

x X 1 
Fig. 90, the shear to the left of the load is — - — , and since this 

JLi 

value under the load is reduced by unity the shear to the right 



of the load is 



X 1. 



INFLUENCE DIAGRAMS 



65 



The shear under the load for any position is given by the inter- 
cept in the triangle. For a system of loads the shear to the left 



& 




Fig. 89. 



Fig. 90. 



of the first load is the sum of the products of the several intercepts 
in the triangle by their respective loads. In the case illustrated, 
Fig. 91, the left reaction or shear to the left of load 1 is 

Ri = Piyi + P 2 y2 + Pzjz + P4V4 + P^ 



ijiiiii® 




Fig. 91. 

If the entire system of loads is moved to the left a distance x, 
but no loads enter or leave the span, Ri increases an amount 

(Pi + P* + Pz + P 4 + P B ) f > 
making the reaction 



Ri' = SPv +SP 



66 



GRAPHICS AND STRUCTURAL DESIGN 

Maximum Shear at any Point 



The vertical shear at a section a distance x from the left 
reaction, Fig. 92, is V Xl = 2Pv. 

Under the load Pi this is reduced by the amount Pi making 
the shear to the right of Pi, V 2 = 2 Py — Pi. If the loads are 
moved to the left until P 2 is at section x from the left reaction the 



, «t> &<$&>$) 


-n i i 1 i 1 

^ — b — 4* — b — H 
Rl < * H 


; 1 

f*2 





Fig. 92. 



shears, providing the same loads only are now on the span, will 
have increased an amount 2P — > making 

F X2 = SPv-Pi + 2p|. 

Comparing V Xl and V Xl it is seen that V Xl will be the greater so 
long as 

D ^ 2P6 Pi ^ 2P 



Under these circumstances the shear at x will be a maximum with 

p 
Pi at x when — - exceeds the sum of all the loads on the span 



P 2P 
divided by that span. If -~ = -— the shears will be the same 

P 2P 

at x with either P x or P 2 at that point, while if -^ < — the 

maximum shear will occur with P2 at x. 



INFLUENCE DIAGRAMS 



6 7 



Had another load P 6 come on the span, Fig. 93, the increase 
in shear, after the load Pi had passed x and advanced a distance 
b to the left of it, would be 

Here 2Pi_5 represents the sum of the loads Pi to P 5 inclusive, but 
not the load P 6 just assumed as coming on the span, while 
2Pi_6 is the sum of the loads from Pi to P 6 inclusive. The dis- 



o CD (t> &> (f) fh 



t 



-f-G->- 

> 



Fig. 93. 

tance c can only range in value from zero to c = b, hence the 
increase in shear will be somewhere between 



m4 



Pi and 



*•«! 



Pi. 



Where the first expression is negative and the latter positive 
both positions of the drivers should be tried. 

Load Pi at section x will give a maximum shear when 

Pi = ZPl-6 

b =- Z 

and load P 2 will give a maximum shear when 

Pi = ZP1-5 
b ^ L ' 

In the case of a uniform load the vertical shear at x is a maxi- 
mum when the portion of the span to the right of x is fully loaded, 
while the shear will be a minimum when the portion to the left 
of x is fully loaded, x being less than one-half the span. 

Position of a system of moving loads to give a maximum shear in 
any panel of a truss with parallel or inclined chords. 



68 



GRAPHICS AND STRUCTURAL DESIGN 



In Figs. 94 and 95, 
SPi is the sum of the loads to the left of the panel. 
2P 2 is the sum of the loads on the panel, here panel 2-3, 
2P 3 is the sum of the loads to the right of the panel. 

m is the number of panels to the left of panel 2-3. 

n is the number of panels in the truss. 

p is the length of a panel. 

Fig. 94. 




--4-J 




Fig. 95. 



The influence lines for loads of unity ab and cd are drawn as 
for shear in a beam or girder. When point 3 is reached the in- 
tercept in the triangle abc is 

1 X {n — m — 1) 
n 



ef- + 



Now as the load moves across panel 2-3 the shear is reduced 
gradually until when point 2 is reached the unit load is deducted 
making the shear 



g h = 



n — m 



m 
n 



The influence diagram for the shear in panel 2-3 is given by the 



INFLUENCE DIAGRAMS 69 

broken line chea. The maximum positive shear in panel 2-3 is 

V = SiVs + 2P 2 ;y 2 - 2P iyi . 
Moving the loads a small distance dx to the left, no load passing 
a panel point or end reaction, the shear becomes 

V + dV = 2P 3 (^3 + dy z ) + 2P 2 (y 2 - dy 2 ) - 2Pi(ji - dyj. 
By subtraction 

dV = XP 3 dy 3 - XP 2 dy 2 + XPidyi. (1) 

Comparing the differential triangles with the similar triangles 
cgh, hegf and fea, we have 

dyi _ _i_ . dy^ _ n — 1 , dyz _ _i_ 

dx np dx np dx np 

Now dividing equation (1) by dx and putting — = o, to ex- 

dx 

amine for a maximum, we have 

dx np np np 

and 

xp _ ZP! + 2P 2 + SP 3 , 

This states that the vertical shear in any panel will be a maximum 
when the load in that panel equals the average panel load for the 
span. It is necessary that a load near the head of the train be 
at the panel point to the right of the panel in which the shear is 
sought. 

Maximum Floor-beam Reaction 

Figure 96 shows two panels pi and p 2 with floor beams a, b and 
c and accompanying stringers. To find the maximum floor- 
beam reaction at b construct the influence diagram, Fig. 96, 
making its ordinate at ef equal one. Let 2Pi be the sum of the 
loads in panel pi and 2P 2 the same for panel p 2 , then for a maxi- 
mum floor-beam load — - = — . Fig. 97 is the influence 

pi pi + pi 

line for the bending moment at b. 



7 o 



GRAPHICS AND STRUCTURAL DESIGN 



Ordinates y and y\ similarly located in Figs. 96 and 97 will be 
to each other as the corresponding intercepts ef and hi or 

2 _ Pi + P2 . 
yi P1P2 
hence 

Plp2 



a OOOO & OOOQO c 





Fig. 96. 

To rind the maximum floor-beam reaction determine the maxi- 
mum bending moment on a beam whose span is / = pi + p 2 at a 
distance pi from the left support a, and multiply it by the sum of 
the panels pi and p 2 and divide this by the product of pi and p 2 ; 
generally pi = p 2 , so that the maximum reaction will ordinarily 
equal twice the maximum bending moment divided by the panel 

width. „ A _, _ 

Moment Table 

^ 1 p i ^ 1 % 1 



Eli 



-a^-l 



«-th 



R 3 



Fig. 98. 

When the stresses are determined by calculation, advantage is 
taken of the following principles in computing. 

The moment due to the loads about the section under P 5 in 
Fig. 98 is 

Mi = P1X1 + P 2 X 2 + P 3 #3 + PiOCi + P5X5' 



INFLUENCE DIAGRAMS 7 1 

If the moment is desired under P 6 , P 6 being a distance c from P 5 , 
then 

M 2 = P 1 (Xi + c) + P 2 (x 2 + c) + . . . P 5 c 

In the second instance if 2P is the sum of all the loads on the 
span to the left of P 6 , then 

M 2 = M l + 2P X c. 

In this way, starting at the head of the train, the moments of the 
loads to the left about a point under each successive load are 
calculated and noted. Having placed the loads in the position 
to give maximum bending at the point to be investigated, first 
find the reaction Ri and then the bending moment at the point 
desired. 

If M 3 is the bending due to loads Pi to P 6 about the section 
under P 7 the bending moment about R 2 of loads Pi to P 7 , inclu- 
sive, will be 

M 4 = M 3 + 2 (Pi - P 7 ) e 

and the reaction is 

where S is the span. 

The bending moment at a section under P 5 then is 
M = Rif- Ml 

The work of calculating stresses due to moving locomotive and 
train loads is facilitated by the use of the following table which 
has been computed in the manner just described for Cooper's 
E-6o loading. 

In this moment table the consecutive wheel loads are numbered 
from the left to the right, and the distances given between 
adjacent wheels, the sum of the distances from the train load 
to each wheel and the sum of the loads from load 18 to and 
including each wheel load are also given. In the body of the 
table above the heavy zigzag line are given the moments of all 
loads between the heavy vertical line at the right of the horizontal 
row of moments and any wheel load to the left, about the vertical 



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INFLUENCE DIAGRAMS 



73 



line. Thus the moment of loads from 8 to 18 inclusive about 
the vertical line marking the beginning of the uniform train 
load is 8,785,500 ft. lbs. (Both pounds and foot pounds are 
expressed in thousands.) 

Below the heavy zigzag line are given the moments of the loads 
to the right of the heavy vertical line in that row; thus the 
moment of the loads 9 to 15, inclusive, about load 9 is 3,720,000 
ft. lbs. Below the body of the table are given the sum of the 
distances from wheel 1 to the several wheel loads, and also the 
sum of the loads from load 1 to any other load including both 
this latter load and load 1. 

A simple example will illustrate one use of this table. In a 
girder whose span is 80 ft. what is the left reaction when load 1 
is over that support? 

Wheel 14 being 79 ft. from wheel 1 when 1 is on the left pier, 
14 will be 1 ft. from the right pier. The moment of wheels 1 
to 14 about wheel 1 is 14,400,000 ft. lbs., read below the zigzag 
line. 

The right reaction is this moment divided by the span or 

n ' = 180,000 lbs. The left reaction is the sum of the 

80 

loads on the span minus the right reaction or 348,000 — 180,000 

= 168,000. The reaction can also be determined by using the 

moments above the zigzag line. The moment of the loads from 

1 to 14 is 13,092,000 ft. lbs., but load 14 being 1 ft. from the 

right pier the moment of the loads as placed upon the girder 

about the right pier is 13,092,000 + the sum of the loads from 

1 to 14 multiplied by 1 ft. 

M = 13,092,000 + (348,000 X 1) = 13,440,000 ft. lbs. 

The left reaction is 

13,440,000 , ,, 

• OJ „ ' = 168,000 lbs., 

80 

the same as found before. 

The use of this table and of the rules just derived by means of 

the influence diagrams will be illustrated by the following problem. 



74 



GRAPHICS AND STRUCTURAL DESIGN 



A through Pratt truss, Fig. 99, carries a train load represented 
by Cooper's E-40 loading.* Find the stresses in the members 
of the second panel from the left pier. 

To find the maximum stress in member EI determine the 
maximum bending at apex 4. The criterion for maximum 
bending at a joint in a truss is that the average load to the left 
of the joint shall equal the average load on the entire span. 

>\ 
"\ 




7 

ISO 



— 1 



Fig. 99. 



Trying load 7 at apex 4: 
( Panel load to left not 
( including load 7 



= Average panel load on bridge. 



103,000 340,000 , , ,, 

-^ = 51,500; ^ — ■ = 5M7o lbs. 

2 

( Panel load to left 

( including load 7 

116,000 340,000 , , ,, 

l = 58,000; i2 ^ — = 56,670 lbs. 

2 6 

To find reaction at the left, take moments of loads about right 
support. When load 7 is at apex 4, which is 50 ft. from the left 
pier, 28 ft. of uniform load will be on the bridge. The moment 
about the right pier then is 

M = 16,364 + (284 X 28) + (28 X 2 X 14) = 25,100 (1000 ft. lbs.) 
from which the reaction at the left is 

„ 25,100,000 c ,•, 

. Ri = -^ l = 1-67,300 lbs. 

15° 

* Note. — Loads and moments will be ffi (f) of Cooper's E-60 loading 
in the moment-tables, page 72; the wheel spacing will remain the same. 



INFLUENCE DIAGRAMS 75 

The bending moment under apex 4 is 

1, (2 =5,100,000 X S°\ ^ r. 11 

M = ( — 2 —J — 2,155,000 = 6,211,700 ft. lbs. 

Here 2,155,000 is the bending due to loads 1 to 6, inclusive, 
about load 7. 
The stress in member EI is 

6,211,700 > ,, 

— — = 207,060 lbs. 

3° 
Stress in member HI. To determine the live load stresses in 
HI, first find the maximum shear in panel 3-4. For the shear in 
any panel to be a maximum the load in that panel must be equal 
to the average panel load on the bridge. 



Load 4 at apex 4. 


Load in panel. 


Average load in panel. 


Load omitting load 4 


Lbs. 
50,000 
70,000 


Lbs. 
302,000/6 = 50,330 
302,000/6 = 50,330 


Load including load 4 





The maximum shear in panel 3-4 occurs with wheel 4 at apex 4. 
To find the reaction under these conditions when load 18 is 9 ft. 
to the left of the right pier we have 

16,364 + (284 X 9) -H (9 X 2 X 4-5) = 19,001 (1000 ft. lbs.) 

^ ,. 10,001,000 , c „ 

Reaction = - 21 = 126,670 lbs. 

150 

The reaction at 3 due to loads in panel 3-4 is 

„ moments of loads 1-4 about 4 
jt = — . 

25 
„ 480,000 „ 

R = ~ — = 19,200 lbs. 

25 
Maximum shear panel 3-4 = 126,670 — 19,200 = 107,470 lbs. 
If tried for load 3 at apex 4, the maximum shear will be found to 
be the same. 
Using this shear the live-load stress in HI is found to be 

HI = 107,470 X ^ = 139,890 lbs. 
30 



CHAPTER VI 
TENSION, COMPRESSION PIECES AND BEAMS 

Designing any piece of a structure requires the determination 
of the resisting forces called forth in it to balance the external 
forces acting upon it. In the case of a purely tension piece this 
requires that the minimum or net section multiplied by the 
allowable working fiber stress shall equal or exceed the total force 
acting in the piece. 

Compression pieces whose lengths do not exceed five times 
their least diameter can be designed by assuming the total load 
equal to the allowable working fiber stress in compression times 
the area of the gross section. In both cases care should be taken 
to have the load distributed over the section as improper applica- 
tion of the external forces to the piece may result in introducing 
bending or injurious local concentrations of stress in it. 

Beams 

Upon the following fundamental conditions of static equilib- 
rium determinations of the required forces or moments acting 
on or in the piece are made. This applies whether the deter- 
minations are made algebraically or graphically. 

The sum of all the vertical forces = o. 

The sum of all the horizontal forces = o. 

The sum of the moments of all forces about any point = o. 

Reactions. — In a beam acted on by several vertical forces 
the sum of the reactions or forces at the supports must equal 
the sum of the loads, and the algebraic sum of the moments 
of all forces and reactions referred to any point must be zero. 
In Fig. ioo 

Pi + P 2 + Pz = Ri + #2. (i) 

76 



TENSION, COMPRESSION PIECES AND BEAMS 



77 



Taking moments about any point in the reaction R 2 we have 

M = Piai + P 2 a 2 + Pza z - RJ, = o. (2) 

The solution of equation (2) gives Ri. R 2 will be given by sub- 
stitution in equation (1). 

Vertical Shear. — The vertical shear at any section is the 
algebraic sum of all the external forces on the left of that section. 
Thus a shear diagram of the beam is shown by Fig. 101. 

At the left support the vertical shear equals the reaction Pi, 
and this value of the shear continues toward the right until 
under the load P 3 the shear becomes Ri — P 3 ; similarly, under 
P 2 the vertical shear is Pi — (P 3 + P2) and under Pi the shear 
is Pi- (P3+P2+P1) = -P2. 



Fig. 100. 
P 3 P 2 I p. 



R, 



-a-j-*J 



m 



J inn 

p 4 



J^ 



Fig. ioi. 



Fig. 102. 



Bending Moment. — The bending moment at any section is 
the algebraic sum of the moments of the external forces on the 
left of that section referred to a point in that section. In the 
beam, Fig. 102, the bending at the section ah is 

M = RiG- P z e - P 2 d. 

Where rolled beams are used generally only the maximum bend- 
ing is required; this will occur where the vertical shear passes 
through zero. (See any book on Mechanics of Materials for 
the proof of this.) In the case illustrated in Fig. 101, the maxi- 
mum bending occurs under load P 2 . Having found the maximum 
bending moment, if the beam is supported laterally, the proper 



78 GRAPHICS AND STRUCTURAL DESIGN 

section can usually be selected from a manufacturer's handbook. 
In building construction beams will commonly be supported 
laterally by flooring, roofing or bracing. 

The external bending moment induces an equal and opposite 
moment in the material of the beam called a resisting moment. 
This is expressed by equation 

e 
where 
M = the external bending moment at a given section, in. lbs. 

— = the resisting moment at the same section. 
e 

f = the extreme unit fiber stress, generally pounds per square 

inch. 
/ = the moment of inertia of the beam section, in inches 4 . 
e = the distance from the neutral axis to the extreme fibers 

in inches. 

The expression - depends entirely upon the form of the beam 
e 

section and is sometimes called the section modulus. 

An example will illustrate the selection of a floor beam. A 
floor beam with a span of 12 ft. is to carry a uniform load of 
14,400 lbs. Select a suitable beam allowing a working fiber 
stress of 16,000 lbs. per sq. in. 

The maximum bending for a supported beam with a uniform 

load is M = —— , where W is the total uniform load in pounds 

and I is the length of the span in inches. 

Substituting the given values in this formula we have 

, T 14,400 X (12 X 12) . „ 

M = ■ J123 :r ■• = 259,200 m. lbs. 

8 

The handbooks usually tabulate the values of - of their sections. 

* For the derivation of this formula see any book on " Mechanics of Materials." 



TENSION, COMPRESSION PIECES AND BEAMS 



79 



/ M 
By transposition of the previously given equation - = —, then 

6 J 



by substitution 



259,200 



= 16.2. 



e 16,000 

The lightest weight standard I beam providing a sufficient sec- 
tion modulus is a 9-in. I beam, weighing 21 lbs. per ft. Its 
section modulus is 18.9. 

Standard Framing 



it 



Lf 




I 



FT 



Ms "Clearance 



9-ie ' Clearance --nj 



Fig. 103. 



A = 



Not infrequently the beam must not only be amply strong 
but it must not deflect excessively under load. When plastered 
ceilings are carried under the beams this deflection is limited to 
¥ J-o of the span. The formula for the deflection at the middle 
of a supported beam carrying a uniform load is 

SWl 3 T 5 WP 

- or / = — 

384.fi/ 3 8 4 £XA 

For steel E = 30,000,000 lbs. per sq. in. 

If the deflection is A = -^— = -7^ = - in., 
360 360 5 

the limiting inertia then will be 

j = 5WI 3 = 5 X 14,400 X 144 3 = 6 
384 E X A 384 X 30,000,000 X I '* 

It is, therefore, evident that the 9-in. I beam weighing 21 lbs. 
per ft. demanded for strength will be amply stiff, its inertia 
being 84.9 or almost twice the inertia required for stiffness. 



8o 



GRAPHICS AND STRUCTURAL DESIGN 



In framing beams into girders, beams or columns each manu- 
facturer has a standard framing. This framing is designed for 
the shortest span and consequently the greatest load for which 
the beam is likely to be used. 



J-i-e- 






18&20 



2- L- 8 4 l 'x i"x Jfe'xlV 
Wt. 37* 






ff" 



I 
I 



]S 



■L'6 4 x4 x %xl 3 
Wt. 31* 



7 



15 




2-L- 8 6Wx %"xO'lo" 
Wt. 27* 



• ! 



2 *5y^£ 






-l-O 



2-L- 66 "x4"x^'xO'7«" 
Wt. 20* 



7, 8,j)_&10 

2^ 



£ 



■2*" 



-l-O o 



2-L's6"x4"x Xl-ad'b" 

Wt. u-# 
Fig. 104. 



, 5 & 6 

2-L *s 6 'x 4x %"x 3" for 6 Wt. 8* 

" 2^" " 5" •• r* 



: • 



-» ||2HT 

f_9_Q 



3"&4" 
2- L's 6"x 4"x Jfjx 2" 
Wt. 6* 



The standard framing, Fig. 104, should not be used for spans 
less than the following: 



7. 


Lb. Span, 
ft. 


I. 


Lb. Span, 
ft. 


I. 


Lb. Span, 
ft. 


I. 


Lb. Span, 


24 

20 
20 

18 


80.O-22.0 
80.O-22.O 
65.O-18.O 
55.O-14.O 


15 
15 

12 
12 


80.0-20.0 
60.0-15.5 
42.O-1 I. O 
40.O-II.5 
31-5" 9-0 


IO 

9 
8 

7 


25.0-9.0 
21 .0-7.0 
18.0-5.5 
15.O-4.O 


6 

5 
4 
3 


12. 25-6.O 
9-75-4-0 
7-5 -3 .0 
5-5 -2.0 



All rivets in standard framing are f in. in diameter. 
As this goes to press the American Bridge Company, in their 
specifications for steel structures, publishes a revised standard 



TENSION, COMPRESSION PIECES AND BEAMS 8 1 

for framing, in which all beams use 4 in. X 4-in. angles from 27 
ins. to 12 ins. inclusive; smaller beams use 6 in. X 4-in. angles. 

When work is being detailed for production in a particular 
shop the standards of that shop should be ascertained and 
adhered to. 

It should be noted in Fig. 103 that to facilitate erection the 
connecting angles extend | in. beyond the end of the web of 
the beam and that the distance back to back of the connecting 
angles is | in. less than the space into which the beam is fitted. 
When drawings are being made for a shop having standard 
framing the beam sketch may be like Fig. 103, no dimensions 
being placed on the standard rivet spaces. 

The advantages of the standard framing are: 

It simplifies shop work, enabling a large number of angles to 
be made at one time. The punching of the rivet holes both in 
beams and angles can be done by means of multiple punches, a 
group of holes being made at one stroke of the punch. The 
drawing-room work is also reduced. The total saving in time 
more than counterbalances any probable waste of material. A 
criticism of the method used for designing these connections is 
that there is twisting introduced into the rivet groups increasing 
the shear on the rivets. Tests made on some full-sized beams 
with standard framing seemed to justify the usual practice as 
being ample notwithstanding the above theoretical criticism. 

When, as is the case for very short spans or for beams with 
thin webs, it is necessary to design special connections the cal- 
culations should be based upon the shearing and bearing value 
of the material; see page 85. 

Riveting. — There are two types of rivet heads: " button 
heads," which approximate hemispherical, and " countersunk," 
which are truncated cones. The button-headed rivets should 
always be used where clearances will permit. Button heads can 
be flattened slightly where additional clearance is needed and 
this requires less work than countersunk rivets. The following 
illustrations, Figs. 105 and 106, give the dimensions of various 



82 



GRAPHICS AND STRUCTURAL DESIGN 



1 





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4 



4 



TENSION, COMPRESSION PIECES AND BEAMS 



83 



rivet sizes and symbols for shop and field driving, and are taken 
from the American Bridge Company's Book of Standards. 

Rivets are designated by their nominal diameter D and the 
length under the head before driving. This length, Figs. 107 
and 108, is made up of the grip, the distance through the material 
held together plus the length called the upset or the stock 
required to form the head and bring the body of the rivet up 
to the size of the rivet hole. It is necessary for the rivet holes 
to be larger than the rivet stock, otherwise there would be 





Fig. 107. 



■*— &rip- -, 
Length *i 

Fig. 108. 



difficulty in placing the rivets in their proper rivet holes and the 
stock would be cold before they could be driven. The holes are 
usually -^ in. larger in diameter than the diameter of the rivets. 
The number of rivets required in any joint depends upon the 
forces being transmitted by the rivets, and upon the rivet values 
in shear and bearing. The value in shear may be that due to 
either single or double shear as the rivet tends to fail by shearing 
in one or two cross sections. The bearing value of the rivet is 
the rivet diameter times the lesser thickness of the materials 
transmitting the forces to the rivets times the allowable unit 
bearing value of the rivet material. The following tables give 
the shearing and bearing values of the usual rivet sizes. 



8 4 



GRAPHICS AND STRUCTURAL DESIGN 



AREA TO BE DEDUCTED FOR VARIOUS SIZES OF RIVETS AND 
THICKNESSES OF RIVETED METAL 



Rivet 
diam. 


Thickness of riveted metal. 


i 3 a 


* 


T B 5 


3 

8 

O.I9 

0.23 
0.28 

o-33 
0.38 
0.42 


0.22 
0.27 

o.33 
0.38 
0.44 
0.49 


I 

0.25 
0.31 
0.38 
0.44 
0.50 
0.56 


T 9 5 

0.28 

o-35 
0.42 
0.49 
0.56 
0.63 


I 

0.31 

o-39 
0.47 

0-55 
0.63 
0.70 


H 

0.34 
0.43 
0.52 
0.60 
0.69 
0.77 


f 

0.38 
0.47 
0.56 
0.66 

0.75 
0.84 


13 
T6 

O.4I 

0.5I 
0.6l 
0.7I 
0.8l 
O.9I 


1 
0.44 

o-55 
0.66 
0.77 
0.88 
0.98 


if 

0.47 

o-59 
0.70 
0.82 
0.94 
1 -OS 


1 


t 

I 

2 

I 
3 

4 

I 

I 


O.09 
0. 12 
O.14 
O.16 
O.19 
0. 21 


O.13 
0. 16 
0. 19 
O. 22 
0.2S 
O.28 


O.I6 
0.20 
0.23 
0.27 
O.3I 

o-35 


0.50 

0.63 

o.75 
0.88 
1.0 
1 -13 



Note. — The size of the rivet hole has been assumed as f inch larger than the nominal diameter 
of the rivet. 



1 Hole Out 



2 Holes Out 




STAGGER OF RIVETS TO MAIN- 
TAIN NET SECTION 



Dimensions in inches. 





f-in. rivet. 


f-in. rivet. 




f-in. rivet. 


f-in. rivet. 




b 


b 




b 


b 


I 


if 


1! 


5 


3ts 


3T5 


I* 


If 


2 


5i 


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2 


2^ 


2\ 


6 


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2j 


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• si 


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7 


3l 


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4 


4 


«tt 


3 


8 


3i 


4l 


4i 


2M 


3i% 


*h 


4 


4i 



GAUGES FOR ANGLES, INCHES 



1 

7" 



*=+ 



H-ffrMH 



Lee: 


8 

Ah 

3 

u 


7 
4 
a* 

3 

1 


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7 

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2 
I* 


if 

1 


1* 

1 


if 
I 


14- 
f 


1 
5 

8 


\ 


£1 


* 






Pq 
























Max. rivet 


7 

8 


7 

8 


7 
8 


3 

4 


! 


1 
2 


3 

8 


I 


! 


I 


£ 



Note. — Special gauges may be used where advisable. 



TENSION, COMPRESSION PIECES AND BEAMS 



85 

















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vO 0_ 10 


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in vq t> oo_ 


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rO 


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l> 


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ro 


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d\ 














pp 












■55 


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OOO 


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r-0 


O* O 


r> in 




10*° 




in Ov ro 






1*3 




t N 






oq 


t^ vO -0 


in 10 




























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ro rj- 


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cs 


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000 
O m 
in N o_ 




hM 


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m 00 
i> ro o_ 




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re 


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jo aJ 


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o_ t£ o_ oo 




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OOOOOO 
cq 00 cq M 
ro fO -q rn <n ■* 


O— H 




rO t? "O 1> 




m"^ 


h" tJ n ui > di 


# g 


| 


tf ro 


00 00 ro ■* 


_« 


w 


•>* ro 00 00 ro •* 




O O 






c 


>o « m w m 

w O O "* O 00 
M w ro tt v© t- 


8 




M Ov 


O ■* O 00 


rt 








ro ■* -O l> 


<U 




u 1 

03 


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6 d 0' d 


<l 


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0066 6 O 




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nj 


10 


m O in O 




03 


IO O IO O IT) O 

t^ <n in t^ 




E 


i> 


N 10 h O 


> 


s 


"8 


ro 10 


O C- 00 O 


ro m vo r- 00 


> . 

a* - 

ctS 



4> 


6 0* 


6 d 0" M 


1 


0" 0* 6 6 w 


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°-o 


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i- s 


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P 














g 






S 


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h 
















fa 















86 



GRAPHICS AND STRUCTURAL DESIGN 



A few examples will illustrate the application of the above 
principles. 

f c = the unit bearing value, pounds per square inch. 

/, = the unit shearing value, pounds per square inch. 
Evidently the rivets in Fig. 109 would tend to shear in but 
one place, along ab, and are, therefore, in single shear. 
ird 2 



Shearing value = :LrL - X/ a . 

4 



Bearing value = d X kXf c - 



]£ 



/ r ^ 



J^X 



l_1 



"O" 



xy 



h\ 





S~\ l/a S~\ /^> 1 ^\ 




a 


* 1 * 


b 


1 


t 1 11 s 


\ 


c 


11 


d 




w t 2 \ w w 1 \ k^> 





o o 



{ 



o o 



o .0 



i 



*l<ti 



Rivet-diam.=d. 



t x < U 



Rivet- diam.=d. 



Fig. i 10. 



Fig. 109. 

In Fig. no the rivets are in double shear as they tend to fail 

along the sections ab and cd. 

ird 2 
Shearing value = — • f 8 . Bearing value = d X hXf c - 

2 

The following numerical example will illustrate the design of a 
joint like this one. 

Example. — A joint of the type shown in Fig. no is to trans- 
mit 15,000 lbs.; h = \ in., t 2 = f in., rivets f in. in diameter, 
the allowable shearing stress 10,000 lbs. per sq. in., allowable 
bearing stress 20,000 lbs. per sq. in. The rivet is in double 
shear while the minimum bearing value will be that of the rivet 
against the |-in. plate as in the other direction two plates each 
f in. thick act. From the tables the rivet value of f-in. 
rivets in double shear at 10,000 lbs. is 4420 X 2 = 8840 lbs. 
The rivet value in bearing on a ^-in. plate is 7500 lbs. As the 
joint is to transmit 15,000 lbs. the number of rivets is found by 
dividing 15,000 lbs. by the smaller of the two rivet values, i.e., 

1 K ,000 

— = 2. 

7500 



TENSION, COMPRESSION PIECES AND BEAMS 



87 



Riveted joints should be designed to avoid eccentric stress if 
possible. Where this is unavoidable such eccentricity should be 
reduced to a minimum and the resultant forces acting upon the 
individual rivets be determined. An instance of such a joint, 
frequently met with, is the connection of the foot of a knee brace 
with the column in a light steel building frame. The force F h 
24,000 lbs., Fig. in, is transferred through the rivets 1, 2 and 3 
to the sketch plate and in turn from the plate by the rivets 4 to 
11, inclusive, to the column angles. The center of gravity of the 
rivet group, 4 to 1 1 , is easily located as its center of symmetry.* 
Rivets 4, 7, 8 and 11 are each 7 J ins. from the center of gravity 
and rivets 5, 6, 9 and 10 are each 6.18 ins. from the same point. 




Fig. hi. 

The rivets must offer a resistance F 2 equal, opposite and par- 
allel to Fi. This resultant force F 2 acts through the center of 
gravity of the rivet group, and the direct force F 2 is shared 
equally by the rivets of the group so that each rivet carries 

~^- — = 3000 lbs. The forces Fi and F 2 are 4.75 ins. apart so 
8 

that the moment opposed by the group is 24,000 X 4.75 = 
114,000 in. lbs. 

The force opposed by a rivet will be proportional to its dis- 
tance from the center of gravity of the rivet group. The rivets 

* When the center of gravity of a rivet group cannot be seen by inspection it 
can be readily calculated. If the group has any axis of symmetry the center of 
gravity must lie on that axis. 



88 GRAPHICS AND STRUCTURAL DESIGN 

4, 7, 8 and n being equally distant from G will oppose equal 
forces and similarly of the rivets 5, 6, 9 and 10. Let S be the 

force acting at 4, 7, 8 and 11, then — — X S is the force acting 

1-7-5° J 
on rivets 5, 6, 9 and 10. The moment of each of the rivets 
4, 7, 8 and 11 will be S X 7.50 and the moment for each of the 

rivets 5, 6, 9 and 10 will be [S X — — J X 6.18. 

V 7.50/ 

The total moment for the eight rivets is 

(4 X 5 X 7-5°) + 14 X S X ^-) = 114,000 in. lbs. 
\ 7.50/ 

30 5 + 20.4 6* = 114,000 in. lbs. 

c 114,000 . „ 

o = — — =2262 m. lbs. 

5°-4 
The force acting on each of the rivets 5, 6, 9 and 10 will be 



(sx 6 -^) = 

\ 7.50/ 



2262 X — — = 1864 lbs. 
7-5o 



The forces F± and F 2 create a clockwise moment so that the 
resisting moment must be counterclockwise and the forces act 
to produce a counterclockwise moment about G. The forces 
act at right angles to the arms or lines joining the rivet centers 
with G. 

The resultant shear on any rivet is found by completing the 
triangle of forces at each rivet, and is indicated in the heavy line. 
This is done on the figure for the 
several rivets and the maximum shear 
comes on rivet 11 and is 5200 lbs., 




1-V* 



while the minimum shear acts on 

rivet 4 and is 1000 lbs. 6 

Should there be no axis of symme- " y f 
try take any two base lines, preferably 

at right angles, and find the distance FlG - II2 - 
of the center of gravity from each of these base lines. Fig. 112 

represents an irregular rivet group. The first base line is 



IY 



TENSION, COMPRESSION PIECES AND BEAMS 



89 



taken, passing through rivets 1 and 5. The rivets are assumed 
as being of the same diameter and the cross section of each rivet 
being A ; then the statical moment about axis x-x will be : 



Rivet. 


Area. 


Arm. 


Moment. 


I 


A 


O 


O 


2 


A 


a 


A .a 


3 


A 


b 


A.b 


4 


A 


c 


A .c 


5 


A 


d 


A .0 



Total area = 5 A . Total Moment = A (a + b + c) . 

The distance g from the axis x-x to the center of gravity will 
equal the statical moment divided by the total area or 

_ A (a -f- b + c) 

It is evident from this expression that the distance g equals the 
sum of the arms a, b and c, etc., divided by the number of rivets. 
In the same way the distance g' from the axis y-y can be found 
and the center of gravity is then located. 

In designing riveted joints the following well-established rules 
are used. (These do not apply to joints in boilers or cylinders 
where maximum efficiencies are desired.) 

1. PREFERABLE MINIMUM DIMENSIONS, INCHES 



Rivet diameters. 


7 

8 


3 

4 


5 

8 


1 
2 


Center to center of rivets * 

Rivet center to sheared edge 

Rivet center to rolled edge 


3 


2§ 


2| 

I 


T 3 
I 
I 



* Sometimes taken three diameters of the rivet. 

2. The maximum pitch in the line of the stress for members 
composed of plates and shapes should be 6 ins. for f-in. rivets; 
6 ins. for f-in. rivets; 4J ins. for f-in. rivets; and 4 ins. for ^-in. 
rivets. Where angles have two gauge lines and the rivets are 
staggered the maximum pitch in each gauge line should be twice 
the above dimensions. 



90 GRAPHICS AND STRUCTURAL DESIGN 

3. Where two or more plates are used in contact the rivets' 
holding them together should not be spaced farther apart than 
12 ins. in either direction. 

4. The pitch of rivets in the direction of the stress should not 
be greater than 6 ins., nor exceed 16 times the thinnest outside 
plate connected, and not more than 50 times that thickness at 
right angles to the stress. 

5. The maximum distance from any edge should be 8 times 
the thickness of the plate. 

6. The diameter of the rivets in any angle carrying calculated 
stresses should not exceed one-fourth of the width of the leg in 
which they are driven. In minor parts rivet diameters may be 
J in. larger. 

7. The pitch of rivets at the ends of built compression mem- 
bers shall not exceed four diameters of the rivets for a length 
equal to one and one-half times the maximum width of the 
member. 

8. Two pieces^ riveted together should always be secured by 
at least two rivets. 

9. Joints with field-driven rivets should have from 25 to 50 
per cent more rivets than would have been required for shop- 
driven rivets. 



CHAPTER VII 
COLUMNS 

Structural engineers generally use either Rankine's formula 
or Johnson's straight-line formula in designing columns. The 
latter is a modification of Rankine's formula and the results 
approximate those given by Rankine's formula within the usual 

working limits of - which will range from 50 to 150. The 

straight-line formula is preferred as the calculations with it are 
simpler. 

The following give these formulae in their usual form: 
Gordon's or Rankine's formula for soft steel, 

/- I5 '°7 - (1) 



1 + 



13,500 r 2 



Gordon's or Rankine's formula for medium steel, 

/- i7 '°°; • « 

1'+— —i 

1 1 ,000 r 

Johnson's straight-line formula for structural steel, 

/= 16,000 - 70-. (3) 

/ = the allowable unit compression on gross section of col- 
umn in pounds per square inch. 
I = the effective length of the column in inches.* 

* The effective length of / will have the following relations to L the total length 
of the column: 

Both ends hinged or butting I = L 

Both ends fixed I = \L 

One end fixed and one end hinged I = \L 

One end fixed and other end free I = 2 L 

9i 



Q2 GRAPHICS AND STRUCTURAL DESIGN 



-4 



r = the least radius of gyration of the column section in inches 

A 

I = the least moment of inertia of the column section in inches 4 . 

A = the area of the column section in square inches. 
Ritter's formula most nearly meets the theoretical require- 
ments but it is not much used as, like Rankine's formula, it is 
cumbersome, and the other formulae are better known. It has 
the advantage of being applicable wherever the modulus of 
elasticity and the strength of the material at the elastic limit are 
known. In the other formulae the constant for the material 
13,500 and 11,000 in the case of Rankine's and 70 in the straight- 
line formula had to be determined experimentally. Ritter's 
formula as generally expressed is: 

'=77% <4) 

The symbols have the same significance as just given for formulae 
(1), (2) and (3). In addition * 

S c = the maximum compressive unit stress desired on the con- 
cave side of the column in pounds per square inch. 
S e = the unit strength of the material at its elastic limit in 

pounds per square inch. 
7r 2 = approximately 10. 

E = modulus of elasticity of the material in pounds per square 
inch. If for mild steel, the following values may be assumed: 
E = 30,000,000, S e =30,000 and S c = 16,000, 

all, pounds per square inch. With these values substituted, 
formula (4) reduces to 

. 16,000 16,000 / >. 

.+ J ' 000 xff = I+ -^-fY s 

10 X 30,000,000 W 10,000 \rj 

In this approximate form it is not especially cumbersome but it 
gives values slightly under those calculated by formula (3). The 



COLUMNS 93 

straight-line formula will be used for all calculations throughout 
the text. 

The American Bridge Company have issued under date of 
Dec. i, 191 2, new specifications for structural steel work. These 
specifications recommend for columns the use of two formulae, 

the first, / = 19,000 — 100 -, to be applied to values of - up to 

120; the second formula applies only to secondary members, 

those permitted to have - values ranging from 1 20 to 200, and is 

/ = 13,000 — 50-- The maximum value of / is placed at 13,000 
lbs. per sq. in. 

Note. — There is probably no section of Mechanics of Materials about 
which students have a less clear conception than that of columns. 

It is here assumed that the student has this fundamental conception 
clearly and thoroughly his own. Such a conception is essential to the 
intelligent design of columns and, in fact, to beams also, as will be shown 
later. Should the student's idea of the subject be in any way vague he is 
urged to review the subject in any Mechanics of Materials with which he 
is familiar. 

Problem. — Plot the values of /given by formulae (1), (2), (3) 

and (5) upon cross-section paper for values of - between 50 and 

r 

150- 

It should be noted in the several formulae that r is the least 
radius of gyration, hence that section will theoretically make the 
best column where the radius of gyration of the section is a con- 
stant for all axes. This requires a circular section, which, for 
economy of material, would generally be a hollow cylinder. On 
this account rolled-steel and cast-iron pipe would make good 
columns, but that cast iron is being but infrequently used in 
important structures while, owing to the difficulty of making con- 
nections to them by brackets or framing, round-steel columns are 
seldom used. 



94 



GRAPHICS AND STRUCTURAL DESIGN 



One of the simplest columns that approaches the theoretically 
ideal section is made by using two channels spaced so as to make 
the radii of gyration of the column section referred to its principal 
axes approximately equal. The columns, Fig. 113, are then 
laced with flat bars riveted to the channel flanges so that the two 
channels are made to act as a unit throughout the column's 
length. 

I\ is the moment of inertia of one channel axis 1-1. 

It is the moment of inertia of one channel axis 2-2. 

a is the area of one channel. A-= 2 a. 




Fig. 113. 

The inertia of the total column section about axis 1-1 is 
h = 2/1. 

Since r = y — it follows that if the radii of gyration are to be 

equal about any two axes of a section the inertias must be equal; 
hence 

Ic = h, 

from which 



y a 



c. 



It should not be understood that columns must always ap- 
proach equal radii of gyration about both axes. In the case of 
columns of comparatively short lengths it may be cheaper, owing 
to the reduction in the work of manufacturing, to use a rolled 
section than a built-up column, even if the material is not used 
to such good advantage. 



COLUMNS 



95 



It therefore frequently happens that single sections, usually 
I beams or angles, make good columns. In some cases the base 
is merely an iron casting of some depth for stiffness, containing 
a pocket cored in it into which the section fits. When the column 
is brought into position some molten soft metal is poured around 
it to hold column and base together. In more important cases, 
angle and plate bases similar to those shown in Figs. 116 and 
117 are used. As previously stated the channel column is one 



i/-%H 



k 



Fig. 114. 



4F 



Fig. 115. 



J 


II II 


f 


1- . -1 


1 1 




Fig. 116. 




Fig. 117. 



of the commonest types. These are illustrated in Fig. 113 and 
the bases for such a column are shown in Figs. 116 and 117. 
The flanges of the channels may be turned in instead of out, 
making a square column, but, owing to the difficulty of rivet- 
ing, it costs more to manufacture. With the flanges turned out 
as in Fig. 113 the lacing may be replaced by plates. Another 
design uses the plate on the outside and is laced on the inside, 
thus permitting of inspection and painting. The sections shown 
in Figs. 114 and 115 are also accessible for painting. In Fig. 
115, the web may be either lacing or a solid plate. This section, 
when deep, makes a good column section to resist combined 



g6 GRAPHICS AND STRUCTURAL DESIGN 

compression and bending. Its strength may be increased by 
riveting flange plates outside the angles. 

A couple of the simplest forms of column bases are shown in 
Figs. 116 and 117. The proper distribution of the load carried 
by the column section to the foundation demands not only ah 
enlargement of the foot of the column but some depth to the 
bed plate. 

Consult paragraphs 90, 92, 96, 115, 126, 128 and 129 of the 
specifications. 

The destruction of the Quebec Bridge due to the failure of a 
compression member has quickened the interest in the safe de- 
signing of columns. A few full-sized tests, but still an insufficient 
number, have been made on latticed columns. See Bulletin 
number 44 of the University of Illinois and tests upon columns by 
Howard, in the A.S.C.E., Vol. 73, with the discussions following 
the article. All the columns under test showed greater stress 
than would be indicated by the usual formulas for column de- 
sign. The suggestions for remedying the difficulty range from 
the suggestion of using fiber stresses on columns only 60 per cent 
of those used in tension to the revision of the column formulas. 
Mr. F. C. Kunz draws attention to the fact that two things, 
ordinarily neglected, may under certain conditions become of 
considerable importance. The first of these is that it is usually 

/' 
considered ample if the-: ratio of one of the members of a column 
r 

between its latticed supports does not exceed the -of the entire 

r 

column. It should be noted that the member may tend to 

buckle between the points of lattice support at the same time 

that the entire column buckles throughout its full length. 

Secondly, although shearing stress may have a negligible effect 

upon columns with solid webs, where the web and flange areas 

are of the usual proportions, and the columns are not too short, 

in other cases it should be considered. 



COLUMNS 97 



Combined Stresses 



In a column the force may act parallel to the axis but eccen- 
tric to it. Here the column is subjected to a bending moment 
P (8 + A) where P is the force acting parallel to the axis and A 
is the maximum deflection due to such eccentric loading. The 
following formula by Merriman gives the maximum resulting 
fiber stress: 



*-5[-+$*^» 



fr = maximum combined fiber stress in pounds per square inch. 

P = eccentric force on column in pounds. 

A = area of the column section in square inches. 

8 = eccentricity of the load P in inches. 

e = distance from the neutral axis to the extreme fibers. 

r = radius of gyration of column section referred to the axis 
about which the bending occurs. 

PP PI 2 



v = 



48 AEr 2 48 EI 



I = length of the column in inches. 
E = modulus of elasticity in pounds per square inch. 

Where a piece is subjected to either compression or tension, 
together with transverse bending, the following formula, due to 
Johnson, is commonly used: 

Me 



fi 



PP 
10 £ 

fb = flexural fiber stress in pounds per square inch. 
M = transverse bending moment in inch pounds. 
e = the distance from the neutral axis to the extreme fibers 

in inches. 
/ = moment of inertia of the section referred to the axis about 

which the bending occurs. 



9 8 



GRAPHICS AND STRUCTURAL DESIGN 



/ = length of the piece in inches. 
E = modulus of elasticity in pounds per square inch. 

The sign (+) is used when P puts the piece in tension, the 
sign ( — ) when compression is produced. 

To this flexural fiber stress must be added the direct com- 
pression ( — ) or the tension (+). The factor given as 10 varies 
with the character of the loads and the ends, being 9.6 for a 
simple beam uniformly loaded, but 12 for a similar beam with a 
central load. Owing to its simplicity this formula is the more 
frequently used. 

When an approximation only is desired the stress due to bend- 
ing may be found by solving the formula/ = — for the extreme 

fiber stress due to bending and adding to it algebraically the 
direct stress due to either tension or compression. This method 
will be satisfactory when the longitudinal tension or compression 
is not large. The total fiber stress should not exceed that 
permitted on the piece. An example will illustrate the use of 
these formulae. 



i 



Fig. 118. 



Fig. 119. 



The member of the upper chord of a bridge is 25 feet long. 
It weighs no lbs. per foot, and has the section shown in Fig. 118. 

The moment of inertia axis 1-1 is 1003.8. P = 238,100 lbs. 

The distance from the top of the upper plate to the center of 
gravity of the built-up section is 6.30 ins. 



COLUMNS 99 

The bending moment due to the dead load is 

,. WL ( no X 25) X 3 00 . „ 

M = — = * ^ *— = 103,125 in. lbs. 

The fiber stress due to combined bending and direct stress as 
given by the approximate method is 

. Me 103, i2<; X 6.30 , „ 
/ = — = — & — 3 ° = 650 lbs. 
J I 1003.8 D 

Total combined stress 

£ , 2^8,100 ,, 
650 + ° ' r = 8100 lbs. 
31.6 

The following result is given by Johnson's formula, 

Me 103,185 X 6.3 

T PP o 238,IOO X 3Q0 2 9? 

I "j I003.8 **-* " 

10 h 10 X 30,000,000 

Total compression 

, . 2^8,100 n ;, 
697 + 6 ' ■ = 8237 lbs. 

The following example will illustrate the application of the 
formulae when the load is parallel to the longitudinal axis of the 
piece but eccentric to it. A 4-in. X 4-in. X iV m - angle 120 ins. 
long has a load of 8800 lbs. applied to one leg. The piece is in 
compression. What is the total extreme fiber stress in com- 
pression? 

/ = 4.97; area = 3.31 sq. ins.; radius of gyration = 1.23. 

By Merriman's formula 

PI 2 8800 X 120 2 

v = = = 0.0177, 

48 EI 48 X 30,000,000 X 4.97 



'r-5E' + e*£fJ] 



5 

8800 r, , 1. 16 x 1. 16 w 1 + 0.0177 

3-3 
Jr = 5300 lbs. 



.31 L 1.23 2 1 - (5 x 0.0177). 



IOO GRAPHICS AND STRUCTURAL DESIGN 

By Johnson's formula 

The fiber stress in bending is 

, Me 



PI 2 
10 E 
, 8800 X 1. 16 X 1. 16 . „ 

fb = — 5— = 26lO lbS. 

J 8800 X 120 2 

4.97 — ■ 

10 X 30,000,000 

The fiber stress due to compression is 
p c = = 2660 lbs. 

Jt = fb +/ c = 2610 + 2660 = 5270 lbs. 
In this case the maximum stress should be kept under that al- 
lowed on the strut as given by a column formula; hence, 

/ = 16,000 — (70 X ) = 5360 lbs. 

\ 0.79/ 

Long Beams Unsupported Laterally 
When a beam of long span is unsupported laterally the upper 
flange being in compression is liable to fail as a column by buck- 
ling sideways. There is no very satisfactory theoretical treat- 
ment of this subject. The compressive fiber stress in the upper 
flange is usually limited by formulae or rules that have been found 
safe. Mr. Christie, basing his conclusions upon tests made on 
full-size beams for the Pencoyd Iron Works, decided it was safe 
to use a desired limiting fiber stress up to a span of twenty times 
the flange width, and that from this point the working fiber 
stress should be uniformly decreased until, at a span of 70 times 
the flange width, the working fiber stress should be one-half the 
maximum desired stress. 

The handbook of the Cambria Steel Company suggests the 
following formula, 

18,000 



A = 

1 + 



3000 ¥ 



COLUMNS 



IOI 



p = allowable compressive fiber stress in pounds per square 

inch. 
I = length of span in inches. 
b = width of flange of beam in inches. 

Mr. C. C. Schneider in his structural specifications limits the 
allowable compression in flanges of girders to 

pi = 16,000 — 200 - when flange is composed of plates. 
pi = 16,000 — 150 - when flange is a channel. 



Ratio of Span to Flange Width. 



100 



x 

S 50 

O 
© 

I 40 

c 
© 

I 30 



10 



i 1 


f 1 




10 

1 i r 






20 




1 r r 


1 r 1 


30 

1 1 1 


, 


1 r 


40 

> i 




r i 1 


t T 1 


50 


















— • 


?i 2w 


































^Sl 


2^e 




^ 


^c^ 


tfHl 






















\ 


^5 


% 






^j 


^ 


V^ 


*i>a 


1 




















^ 


> 


^yS? 


^ 


><?^ 






«tf 
























s^ 


^ 


%j 


^"V^ 


t 


<x 






























<& 


^ 


















































Curves give the $ of the desired Maximum. 

Fiber Stress for use in Designing Long 

Beams unsupported Laterally. 





























































































10 20 30 40 50 60 70 80 90 100 110 .120 130 140 150 160 170 
Ratio of.Span to Radius of Gyration of Compression Flange 

Fig. 120. 



In this case the fiber stress is given somewhat lower than for 
beams, which is consistent, as the shallower the beam the more 
the lower flange, which is in tension, reinforces the compression 
flange. 

The 191 2 specification of the American Bridge Company limits 



102 GRAPHICS AND STRUCTURAL DESIGN 

the span unsupported laterally to forty times the flange width, 
and when such span exceeds ten flange widths the fiber stress is 

to be reduced to that given by the formula 19,000 — 300 -• 

The preceding curves, Fig. 120, are based on formulae derived 
by the author in an attempt to explain the necessity for re- 
ducing the fiber stress in beams where the beam was unsupported 
laterally. The derivation of these formulae is fully explained in 
the Proceedings of the Engineers' Club of Philadelphia, for April, 
1909. 



CHAPTER VIII 

GIRDERS FOR CONVEYORS 

The half plan and section of this girder are given in Fig. 121, 
the elevation in Fig. 122 and the end view in Fig. 123. 
Assumed Loading. — 

Metal at 105 lbs. per lineal foot 5*360 lbs. 

Corrugated steel covering 2,000 lbs. 

Foot-walk and sheathing 2,420 lbs. 

Snow 6,120 lbs. 

Total dead load 15,900 lbs. 

Conveyor load: 

Uniform load on conveyor 3,900 lbs. 

Weight of conveyor 3,700 lbs. 

Total moving load on conveyor . . 7,600 lbs. 
Allowing 25 per cent additional for impact 1,900 lbs. 

Total 9,500 lbs. 

As the conveyor loading is not carried equally by both girders 
the maximum load on either girder must be calculated. 
The total maximum load on one girder 7950 + 6500 = 14,450. 

The apex load is 5 ~ 2050 lbs. 

Wind Load. — The horizontal wind load is assumed at 20 lbs. 
per sq. ft. of vertical projection. The apex wind load on the 
upper horizontal girder is 20 X 4 X 7! ~ 590 lbs. 

The stresses must now be determined for the truss under the 
given loading. This may be done graphically or the stresses 
may be found very readily by the method of coefficients . To 
illustrate the procedure both methods will be used. 

The method of coefficients will be used first. For the explana- 
tion of this method see page 55. These calculations are made 

103 



io4 



GRAPHICS AND STRUCTURAL DESIGN 





T 




i 


— 





3l2"V u K \R X „S-' 



" r 


i7 T~~ '" 




1 L 




4 








« m ' > 
























*■* >^>^ 
















\> 


r 


\ 4 




n|^ 


1 \ 














^fei- 


kl 


u 




^„K x, g xg- 




g; P, 












^^ 




i $ 




J ^. 




% * 








3 — 




t« N 




s s 






S\ j, 


3 d 








1 -° 




J^ 


[ j£ 


J /° 




; \~" 


14 




^ 


\v 


VI 




<* 






El 




U 






*■:> 


1 i 


[ 














J E? ! 








^ 


j 






Ml 


n 1 






y ' 


- 4 






■j 1 




y 


* i .. 


j — 


§\\ 



4750* 



{■4750^ 



GIRDERS FOR CONVEYORS 

i,, 4750# 



Ri 



6500* 



V^ L 



Fig. 124. 



Fig. 125. 



I05 











Wtangfl 








- 


-5 


-6 




/ 


\ H 


\ J 








\ 1 


seed 


/, 


\+a 

1 \ 

G \ 


\ +1 
1 \ 








\ 


< 


+•3 


+3 


+5 


+6 










D 





' B 1 


r A 1 


■ 


r 1 


\ 




3W 


Wtai 


*sd 












DEAD LOAD 
AND LIVE LOAD 
DIAGRAMS. 



J H 



E-A 



Fig. 127. 



for Fig. 125. The graphical solution for the main truss is made 
in Figs. 126 and 127. 

The wind bracing and graphical analysis are given in Figs. 128 
and 129. 



io6 



GRAPHICS AND STRUCTURAL DESIGN 



6 = 45° tang 45 = 1 sec. 45 = 1.41 

Stress EF = — 3 X 2050 X 1.41 = — 8670 lbs. 

Stress EH = — 5 X 2050 Xi=- 10,250 lbs. 

Stress EJ = EK = — 6 X 2050 X 1 = — 12,300 lbs. 

Stress FD = + 3 X 2050 X 1 = + 6150 lbs. = stress GC. 

Stress IB = + 5 X 2050 X 1 = + 10,250 lbs. 

Stress KA =+ 6 X 2050 X 1 =+ 12,300 lbs. 

Stress GH =+ 2 X 2050 X 1.41 =+5780 lbs. 

Stress // =+ 1 X 2050 X 1.41 =+ 2890 lbs. 




590 



E-F-A 
B 
C 
D 

Fig. 129. 
Selection of Members. — The upper chord. 

Maximum compressive stress, dead and live load 12,300 lbs. 

Maximum compressive stress, wind load 3,54° lbs. 





Fig. 128. 

H 


/ 




/ 


K 


zq 


/ 




G 


/ 



Total 15,840 lbs. 

Try two 3 in. X 2 in. X J-in. angles spaced \ in. back to back, 
with their long legs parallel. r 2 = 0.84. Their length is 88 ins. 

I _ SS 
r 0.84 



= 105. 



is 



The allowable fiber stress according to the straight-line formula 
/1 = 16,000 — 70 - = 16,000 — (70 X 105) = 8650 lbs. 



The load these angles will carry is 2 X 1.19 X 8650 = 20,600 lbs. 



GIRDERS FOR CONVEYORS 107 

End Strut EF, Fig. 125. — Its length is 120 ins. and its load 
8670 lbs. Try one 4 in. X 4-in. angle. Its radius of gyration 

about its diagonal axis is 0.79. - = =152. The allowable 

fiber stress according to the straight-line formula is 
/1 = 16,000 — (70 X 152) = 5360 lbs. 

The bending moment due to the eccentric loading, when the 
angle is assumed as -j 7 g in. thick, is 8670 X 1.16 = 10,060 
in. lbs. 

The fiber stress resulting from combined bending and com- 
pression is given by the formula 

Me 



f = 



PI 2 
10 E 



see page 97. 



, 10,060 X 1. 16 , „ 

/ = —Z7 ;— = 2560 lbs. 

8670 X 120 2 ° 



4.97 



10 X 30,000,000 



The allowable stress according to the straight-line formula 
being 5360 lbs., deducting 2560 lbs. leaves 2800 lbs. per sq. in. 
The total allowable direct stress on the angle then is 3.31 X 2800 
= 9270 lbs.; this, exceeding the 8670 lbs. acting on it, the angle 
is satisfactory. 

Lower Chord. — The maximum stress is 15,840 lbs. Trying 
one 4 in. X 3 in. X jq in., and placing the long leg horizontally, 
its inertia about its axis parallel to the short leg is 3.38; 
the distance from the back of the short leg to the axis is 
1.26 ins. 

The bending moment 15,840 X 1.26 = 19,960 in. lbs. 

The fiber stress due to bending is 

/. Me = — i 9 , 9 6oxi. 2 6 

' I+ 1L 3 . 3 8 + ^ oX882 

10 rL 10 X 30,000,000 



108 GRAPHICS AND STRUCTURAL DESIGN 

The allowable stress on the net section is/ = 16,000 — 6650 = 
9350 lbs. 

The total allowable force on the piece then is (2.09 — 0.27) X 
9350 = 17,020 lbs. 

Vertical HI. — This is a compression piece carrying 2050 lbs. 
One 3 in. X 3 in. X J-in. angle will be tried. 

The bending due to eccentric loading is M = 2050 X 0.84 = 
1720 in. lbs. 

The fiber stress due to this bending is 

, Me 1720 X 0.84 „ 

J ~ ^r = • -^-zrz — = 1220 lbs. 

J T PI 2 2050 X 88 2 

I 1.24 — 

10 E 10 X 30,000,000 

The allowable compression, according to the straight-line 
formula, is 

/ = 16,000 — 70 - = 16,000 — (70 X ) = 5600 lbs. 

r V 0.59/ 

The net compression allowed per square inch of the section is 
5600 — 1220 = 4380 lbs. 

The area of the angle multiplied by 4380 is 1.44 X 4380 = 
6300 lbs. Hence, the 3 in. X 3 in. X |-in. angle is more than 
ample. 

Diagonals. — The maximum tension in the diagonals is 5780 
lbs. Try one 2§ in. X 2 in. X J-in. angle. Its inertia about an 
axis through its center of gravity and parallel to its long leg is 
0.37. The distance from the back of the angle to the above 
axis is 0.54 in. 

The bending moment due to the eccentric application of the 
load to the angle is 5780 X 0.54 = 3120 in. lbs. 

The extreme fiber stress due to this bending is 

, Me 3120 X 0.54 „ 

f = 7~w = — , Msoxitt* =2590lbs ' 

/ -| ■ 0.37 H — 

10 £ 10 X 30,000,000 

The allowable stress per square inch of the net section is 16,000 
— 2590 = 13,410 lbs. Net section X 13,410 = (1.07 — 0.22) X 



GIRDERS FOR CONVEYORS 



109 



13,410 = 11,400 lbs. This is satisfactory, being the smallest 
angle we can use. 

Wind Bracing. — The maximum stress in the diagonals is 
2550 lbs. The smallest flat allowed being 2 X I in., and as this 
will carry [(2 X 0.25) — 0.22] X 16,000 = 4480 lbs. (the net area 
times the allowable fiber stress), it is ample. 



Strut in Upper Horizontal Bracing 

£ 2 = 500X45 ^ 400lbs . 
57 
The bending moment under the load between R x and R 2 is 
M = 400 X 12 = 4800 in. lbs. 



JtfjRidg* Pole 






*F 



-< 33— J 

R 2 



— 57^—" 



Fig. 130. 



30 — 

Fig. 131. 



Trying 3 in. X 3 in. X J-in. angles, their inertia about an axis 
through their center of gravity and parallel to a leg is 1.24, the 
distance from the back of the angle to this axis is 0.84 in., the 
radius of gyration referred to this axis is 0.93, while the radius 
of gyration referred to the diagonal axis is 0.59. 



I 



for the full length 



90 
o-93 



- for the length of < 7 ins. = ~^- 
r 0.59 



97- 



The allowable stress / = 16,000 — ( 70 X - ) = 16,000 — (70 X 97) 
= 9210 lbs. 

The fiber stresses due to the bending of 4800 in. lbs. in 57-in. 
span are, 



, _ Me _ 4800 X 0.84 

I 1.24 



1U r 4800 X 2.16 „ 

3250 lbs. f e = * = 8350 lbs. 

1.24 



HO GRAPHICS AND STRUCTURAL DESIGN 

The bending due to the eccentric loading as a column on 
member GH, Fig. 128, will produce a fiber stress of 

/. - M \ n - (" 8o x °-y x Q-84 _ jas _ 68o lbs , 

T PI 2 1 180 X 57 2 1.23 

I 1.24 2* ° 

10 E 10 X 30,000,000 

, 680 X 2.16 „ 

f* = ~~^ = I75 ° lbs ' 

0.04 

The direct compression = force divided by the area of the 

section = = 820 lbs. 

1.44 

The maximum compression will be 8490 lbs., which, being 
under the 9210 lbs. allowed the section, is satisfactory. 

The beams serving as struts in the lower horizontal bracing 
carry a total load of 1080 lbs., 

^-^- + 400 (snow load) = 1080 lbs. 

7 

R 2 = Io8 ° X 57 = 685 lbs. 
90 

M = 685 X 33 = 22,605 in. lbs. 

Assuming the channel unsupported laterally for its length of 
88 ins., and trying a 6-in. channel, 

88 
length -r- flange width = = 45- 

Allowable fiber stress about 11,000 lbs. per sq. in., 

M I 22,605 , 

— = - = — 1 — a = 2.06. 
/ e 11,000 

This suggests about a 5-in. channel. 

The direct compression being n 80 lbs. and applied at the 
upper flange the moment due to eccentric loading is M = 
1180 X 2.5 = 2950 lbs. 

, Me 2950 X 2.5 „ 

' = — W nsoxss* =Iooolbs - 

I 7.4 

10 E 10 X 30,000,000 



GIRDERS FOR CONVEYORS III 

Direct stress = 605 lbs. 

The extreme compressive stress due to bending of a 5-in. U is 

. Me 22.601; ,, 

/ = — - = — J — * = 7535 lbs. per sq. in. 
1 3.0 

Total compressive stress 7535 + 1000 + 605 = 9140 lbs., being 
below 11,000, is satisfactory. 



CHAPTER IX 

TRUSSES, BENTS AND TOWERS 

It is frequently necessary around works to carry large gas or 
air mains overhead. These mains are sometimes lined with brick 
to prevent radiation loss from the hot gas or air. There is 
consequently considerable load per foot, and as long spans are 
preferable, thus avoiding too large a number of posts or bents 




Fig. 132. 

that would block up the yard, the pipe must be trussed. Fig. 
132 represents such a trussed pipe. Care should be taken that 
any load transferred to or from the pipe should be distributed 
over the pipe by a saddle and if necessary the pipe should be 
reinforced at such points. In Fig. 132, AB is the load in the 
vertical member, "D," AC the compression in the pipe, and BC 
the tension in the rods. 

The span over which the pipe section acts as a beam is -* 

Steam, air, gas and water pipes may be carried upon a light 

112 



TRUSSES, BENTS AND TOWERS 



113 



trussed bridge somewhat resembling the conveyor truss (see 
page 104), or may be hung from a cable, as shown in Fig. 133. 

The stress in the wire rope depends upon the load carried and 
upon the sag permitted in it. As the weight of the rope will be 
small compared with the load, the effect of the weight of the 
rope may be neglected. In Fig. 133, having assumed the rope 
diameter, lay off R representing the maximum desired pull on 
the rope, then draw the horizontal and vertical components H 
and V. Taking the forces acting about apex 1, the horizontal 

Fig. 133. 




Fig. 134. 



component of BC must equal the horizontal component of CA. 

Similarly, the horizontal components of the forces acting at 

W W 

apex 2 are equal. Now in Fig. 134, XD = — and DA = — 

8 4 

E is the horizontal component of the force BC acting in the 

guy rope. CA and CD give the magnitudes and directions of 

the rope stresses and the sag of the rope may be found by drawing 

the strings in Fig. 133, corresponding to these rays in Fig. 134. 

Should the sag prove objectionable a heavier rope with 

greater stress or a greater stress may be used in the first rope if 

permissible. 

W 
The vertical load on the post, Fig. 133, is V -\ 



H4 



GRAPHICS AND STRUCTURAL DESIGN 



Bents, Fig. 135, may be used to support both pipes and wires. 
If the bent is an end one and guyed the vertical load will be 
determined as in the preceding example. If it is an intermediate 
bent it will not carry the vertical component of the guy rope but 
merely its proportion of the weight of the wires and pipes. It 
may also be subjected to wind load, in which event the stresses 
in the bracing can be determined and these pieces selected for 
their respective loads, see Fig. 137. The wind acting on cylin- 
drical surfaces is commonly assumed as t 6 q- of the pressure that 
would act on the vertical projection of that surface. 




WIND 

LOAD 

STRESS 

DIAGRAM 



P-Q 



DEAD 

LOAD 

STRESS 

DIAGRAM 



Fig. 135 



Fig. 138. 



As the loads are generally very light and not subjected to 

shock, the values of - are permitted to reach 180 or 200. 
r 

As the wind load is transmitted to the truss proper by pieces 
subjected to bending, the diagram, Fig. 136, has been started by 
connecting the wind force to the truss through the dotted or 
substituted members. The dead-load stress diagram is shown 
in Fig. 138. 

Towers for Transmission Lines. — There is no standard 
practice but the four-angle type is the most common. The 



TRUSSES, BENTS AND TOWERS 115 

loading upon these towers consists of dead load of towers and 
weight of wire, also ice, which may add materially to this direct 
load in localities visited by sleet. A live load due to the wind 
acts at right angles to the dead load, and in addition there may be 
a direct pull along the line caused by a wire or wires breaking. 
The ice load is generally assumed as a §-in. coating around the 
wire. 

Wind loads assumed as acting on wires vary considerably. 
Mr. R. Fleming in an article in Engineering News, Nov. 28, 191 2, 
recommends 30 lbs. per sq. ft. of exposed surface upon the tower, 
15 lbs. per sq. ft. of projected area of the bare wire, reduced to 
10 lbs. for ice-coated wires. He further recommends that poles 
carrying three wires be designed to resist the unbalanced pull, due 
to one wire breaking, while for six wires two of them shall be 
considered as breaking. The unbalanced load is the force that 
breaks the wire and the following percentages of the ultimate 
strengths of the wires are recommended : the full ultimate strength 
up to and including a No. 4 wire; a No. 3 wire, 90 per cent; a 
No. 2 wire, 80 per cent; a No. 1 wire, 70 per cent; a No. o wire, 
60 per cent; and for all larger wires, 50 per cent. The ultimate 
strength per square inch of hard-drawn copper wire is from 
50,000 to 65,000 lbs. The towers are designed to meet the 
requirements of the worst combination of these conditions. 

Using ordinary structural steel, Mr. Fleming recommends the 
following formulae for columns : 

f c = ^-« — , for the most exacting city service. 

1 + 



18,000 r 2 



; 27,000, , . 

j c = ■ , for less exacting service. 

1 + 



10,000 r 



In estimating the loading upon the poles the additional load- 
ing due to the line running on an incline or rounding a curve 
should not be overlooked. This is generally provided for by 



Il6 GRAPHICS AND STRUCTURAL DESIGN 

placing the poles closer together in these places rather than 
changing the tower design. 

In estimating the tension in the line and length of the wire 
between supports it is common practice to assume the curve of 
the line a parabola which differs but slightly from the catenary 
and is much more readily computed. 

5* = distance between poles in feet. 

L = length of wire in feet, measured along the curve. 

d = sag or drop in wire in feet, measured midway between 
poles. 

H = tension in lowest point of curve in pounds. 
c = tension constant. 

w = resultant load, measured in pounds per foot of wire. In 
winter the weight of wire and ice acting vertically 
and the wind acting horizontally. In summer the 
weight of wire acting vertically and the wind on the 
bare wire acting horizontally. 

L-S + £, (xj L = S + f s , (3) 

For a more complete mathematical discussion of this subject, 
see Bulletin No. 54 of the University of Illinois. " Mechanical 
Stresses in Transmission Lines," by A. Guell. Jan. 22, 1912. 

The length of the wire at the summer temperature will increase 
due to the temperature but will have less extension due to the 
loading, as the vertical load will not include the weight of the ice, 
and the wind will not act on so great a surface. Mr. Guell gives 
the following formula: 

Ll = S + — + aLt- ^ > . ( 5 ) 

Note. — This discussion applies only to spans where the points of at- 
tachment to the poles are at the same elevation. The pull in the wire at 

the pole is T = UH 2 + — lbs. 



TRUSSES, BENTS AND TOWERS 



117 




SECTION A- A 



// 1 '/ 
12 x % Plate 



, 'HI. Ground Line 

1% Bolt 

Top upset to 2}4" 






Fig. 139. 



Il8 GRAPHICS AND STRUCTURAL DESIGN 

Here 

Zi = length of the wire along the curve, measured in feet, 

at the summer temperature. 
L = length of the wire for winter temperature and loading. 
a = coefficient of expansion per degree Fahr. = 0.00000956. 
/ = temperature range, degrees Fahr. 
E = tension at low point in curve in winter, pounds. 
Hi = tension at low point in curve in summer, pounds. 
A = area of wire, square inches. 

E = modulus of elasticity, pounds per square inch. For 
hard-drawn copper wire, 12,000,000 to 16,000,000 lbs. 
per sq. in. 

The stresses in the wire will be lower in summer than in winter, 
but it is necessary to estimate the sag for the summer. It may 
be approximated as follows. In equation (5) estimate the value 
of L h omitting the last term, as Ei is not known. A preHminary 
estimate of Ei may be made by assuming 

Hi = m 

E w 
Here 

Wi = resultant load per foot in summer. 
w = resultant load per foot in winter. 

This value of Ei would then be substituted in equation (5) and 
the last term need not be discarded. Having now found L\ 
substitute this value for L in equation (1) and solve for c. This 
value of c used in equation (2) gives an approximate value of d. 
Now in equation (4), w is known from the loading and E can be 
found. This is an approximation to E\ required in equation (5) 
and the value of L\ may now be revised by placing this value 
of Ei in this equation. The trials may be repeated until the 
desired degree of approximation is reached. 

Cross arms should be designed for a minimum vertical load 
of from 1000 to 1200 lbs., also for an unbalanced horizontal 
pull due to the wires breaking and estimated as previously given. 



TRUSSES, BENTS AND TOWERS 



II 9 



The twisting of the tower due to such unbalanced loading should 
not be overlooked. 





• • 




• • 


— 






16" 



















Concrete 



14 



■d?2* 



-loO- 



FlG. 140. 

Figure 139 shows one type of tower and foundation and follows 
a design by Westinghouse, Church, Kerr & Co., as does also 
Fig. 140, which illustrates a type of foundation with greater 
base area. 



CHAPTER X 

DESIGN OF A STEEL MILL BUILDING 

The design of a steel mill building as generally carried out 
requires considerable experience and judgment on the part of 
the designer. The stresses in simple trusses carried on walls are 
usually determined for an equivalent load rather than by the 
separate determination of the dead-, snow- and wind-load stresses 
and their combination for maximum effect. The following 
equivalent loads may be used on spans under 80 ft. 



Kind of roof. 



Lbs. per sq. ft. 
exposed surface. 



Gravel or composition roofing on boards 

Gravel or composition roofing on 3-in. concrete . 

Corrugated steel sheeting 

Slate on boards 

Slate on 3-in. concrete, flat 



45-So 
60 
40 
50 
65 



Where no snow need be considered these figures can be re- 
duced by 10 lbs., excepting that no roof shall be designed for less 
than 40 lbs. per sq. ft. 

Stiffening the Structure. — The structure is stiffened first 
where possible by introducing a knee brace, running from a 
panel point in the lower or upper chord of the truss to a point as 
low as convenient on the column. Where the column carries a 
crane runway this knee brace must generally be omitted and is 
usually replaced by a knuckle brace which stiffens the connection 
between the column and the truss. 

The structure is also braced in the plane of the lower chords of 
the truss to hold the tops of the columns at constant distances 
apart, and in some cases to carry the wind forces acting along 
the side of the building to the transverse bracing at its ends. 

Note. — See Specifications § 66 to § 135. 



DESIGN OF A STEEL MILL BUILDING 121 

To 'prevent the trusses rotating about their lower chords, 
bracing is placed in the plane of the .upper chords. This bracing 
is commonly omitted, in alternate bays. 

To resist the wind pressure on the ends of the buildings and 
the cumulative pressure along the roof, together with any crane 
thrust, lateral bracing is placed in the plane of the building 
columns. This bracing usually includes an eave strut which 
is frequently a light latticed girder composed of four angles, the 
web being composed of diagonal lattice bars. 

Where possible the diagonal bracing should extend to the ends 
of the columns. Where the column height is considerable an 
additional strut is placed about half way up the column. Both 
eave strut and intermediate strut run the entire length of the 
building and fasten to each column. The diagonal braces are 
commor^y placed at the end bays and then at intervals along 
the building as deemed necessary by the designer. Where the 
bracing cannot be extended to the column bases the columns 
must be designed to resist the bending due to the horizontal 
forces resulting from the wind acting on the end of the building 
and any crane thrust. This bending will be influenced by the 
way the column bases are secured and may be treated similarly 
to the transverse bent, page 45. 

The transverse bracing in the ends of the building is intended 
to resist whatever of the forces acting on the side of the building 
from wind pressure, or thrust on the columns from the crane, 
may be transferred to the ends by the structure. 

In the simpler buildings it is possible to assume certain exter- 
nal forces resisted by a particular bracing truss and then make 
the design accordingly. In the more complicated buildings the 
designer modifies his calculations by what has been previously 
found to be satisfactory. 

The following sketches illustrate one general plan of steel mill 
bracing, Figs. 141 to 143. See also Figs. 156 to 158. 

A. number of types of roof trusses are shown in Figs. 144 
to 151. 



122 



GRAPHICS AND STRUCTURAL DESIGN 




TRANSVERSE 
BENT 



Fig. 141. 



ELEVATION 



Strut 



V 



V 



;&ir; 



-g£ 



X 




Eave Strut 

PLAN 
Fig. 143. 



DESIGN OF A STEEL MILL BUILDING 



123 



The number of panels into which the upper chord should be 
divided will depend upon the span of the truss and upon the roof 
covering or sheathing. In the case of corrugated-steel roofing, 
it is desirable that the sheets should extend over three purlins. 
That the roof covering be amply stiff it is necessary with the 
ordinary gauges to limit the distance between purlins to from 





\Fig. 144. 



Fig. 145. 





Fig. 146. 



Fig. 147. 




Fig. 148. 



Fig. 149. 





Fig. 150. 



Fig. 151. 



4 to 6 f t. ; consequently, the purlins are generally spaced about 
4 ft. In the case of other roof coverings stiffness is also re- 
quired, but as most of these coverings are laid upon sheathing 
the purlins can be spaced farther apart providing the sheathing 
is made correspondingly thicker. If the upper chord is not in- 
tended to take bending it is necessary to have a member of the 
truss connected to the upper chord under each purlin. 



124 



GRAPHICS AND STRUCTURAL DESIGN 



Frequently, however, the upper chord is designed of a plate 
and two angles forming a tee-section. In this case the section is 
calculated to resist bending and the purlins can, therefore, be 
spaced as desired, and a less complicated truss used. 

The pitch of the roof will vary with the character of the roof 
covering. The following table gives the usual pitches. 



Kind of Roofing. 



Minimum pitch. 



Usual pitch. 



Corrugated steel, 

Slate or tile 

Shingles 



Gravel 

Asphalt 

Patent compositions . 



Maximum pitch 



m 12 



Determination of Stresses. — The methods of combining the 
stresses due to the several classes of loads vary with the de- 
signer. In the Engineering News of Feb. 4, 191 5, Mr. R. Flem- 
ing of the American Bridge Co. recommends the following pro- 
cedure. 

In the design of simple trusses the stresses allowed to resist 
wind loadings may be increased 50 per cent over those used to 
resist ordinary live and dead load stresses, thus making a wind 
load of 15 lbs. equivalent to 10 lbs. using the working stresses 
for other loads. The snow load per square foot horizontally 
will range from 20 lbs. in New York to 30 lbs. in parts of New 
England. On a roof with 6-in. pitch this will range from 16.6 to 
25 lbs. per sq. ft. over the entire surface. For the combined 
wind and snow, a load of 25 lbs. per sq. ft., acting vertically over 
the entire surface, is ample for roofs in the latitude of New York 
City. To this should be added the weight of the trusses, purlins 
and roof covering, reduced to the square foot of roof surface. 
The total load for which a roof should be designed, however, 
should not be less than 40 lbs. per sq. ft. of exposed surface, ex- 
cepting in tropical climates with no snow where 30 lbs. per sq. ft. 
may be the minimum, or where snows are severe the minimum 



Fig. 157. 




4Ls3of2KxK ^ 
CJ Double Lacing 2% \/\ t 

lL6ft"x8&*x>£ 



Fig. 158 



DESIGN OF STEEL 
MILL-BUILDING 




DESIGN OF A STEEL MILL BUILDING 125 

load should be increased to from 45 to 50 lbs. per sq. ft. of 
exposed surface. In the present state of our information 
he does not recommend an allowance for suction. The pos- 
sibility of stress reversal should be met by using only stiff sec- 
tions. This eliminates the use of all rounds and flats in roof 
trusses. 

Knee-braced Bents. — In the case of knee-braced bents, Mr. 
Fleming recommends first proportioning the truss members for 
the stresses due to a total uniform load, using 16,000 lbs. per 
sq. in. net tension and the same reduced by formula for gross 
compression. 

If the wind stress in any member, as given by the wind load 
stress diagram, similar to Fig. 158b, exceeds the stresses due to 
a uniform loading of 10 lbs. per sq. ft. applied vertically, that 
member is proportioned for the maximum wind stress plus the 
stresses from the uniform loads other than wind, using a working 
stress exceeding that in the first calculation by 50 per cent. 
In no case is a smaller section than that first chosen to be used. 
Members liable to a reversal of stress may be designed of two 
angles instead of one where one would have served for the ten- 
sion. Knee-braces being subjected to wind stress only, he de- 
signs with the larger working stress. 

Columns. — These are first designed to carry the direct stress 
due to the total uniform load from the truss, noting the flange 
area required. From the maximum bending moment due to 
the wind, the forces being taken from the wind-load stress 
diagram, see Fig. 158b, the sectional area required for the 
flanges is found, using the larger working stresses and consider- 
ing the column as a beam. If tins flange area is not more than 
one-third of that first found no change is made, if it is more than 
one-third the excess is added to the previously found flange 
area. The compressive stress due to the overturning (the verti- 

Note. — Articles upon Wind Pressures and Wind Stresses in Structures 
by Mr. R. Fleming appeared in the Engineering News of Jan. 28; Feb. 4, 
11 and 25; and March 11 , 191 5. 



126 



GRAPHICS AND STRUCTURAL DESIGN 




DESIGN OF A STEEL MILL BUILDING 



127 



cal component of the wind reaction R 2 in Fig. 158a) need not be 
considered unless it exceeds the stress from the wind portion of 
the uniform roof load. 

Design of a Steel Mill Building, Figs. 152 to 158. — The truss, 
Fig. 159, will be designed according to the specifications para- 




Fig. 159. 

graphs numbers 66 to 135 inclusive. The span is 86 ft. and the 

rise 19.22 ft., thus making the angle of slope 24 degrees 5 minutes. 

The several members have the following calculated lengths, 

CH, DI, EK, FN, GO, 9.42 feet. 

HI, 4.21 feet. LM, 12.63 ft - JK , 8 -4 2 ft - 

//, HA, J A, LA, 10.32 ft. MP, OP, NM, 11.34 ft. 

NO, 6.31 ft. Length of rafter, 47.1 ft. 

The normal wind pressure by Duchemin's formula is 14 lbs. 
per sq. ft. of roof surface. 

The dead load is estimated at the following pounds, per square 
foot of horizontal projection of roof, trusses, 3.7 ; purlins and brac- 
ing, 2.8; roof covering, 3.5, making a total of 10.0 lbs. per sq. ft. 

The trusses are spaced 17 ft. center to center. 

The apex dead loads are (86 X 17 X 10) -f- 10 = 1462, say 
1460 lbs. 

The apex wind load normal to the roof is 14 X 9.42 X 17 = 
2240 lbs. 

The apex snow load being due to a load of 20 lbs. per sq. ft. 
of horizontal roof projection will be twice the apex dead load, 



128 GRAPHICS AND STRUCTURAL DESIGN 

and the stresses due to the maximum snow load may be taken 
as twice those due to the dead load. 

The dead load stress diagram is given in Fig. 153. The 
wind load stress diagram is shown in Fig. 158b. In this diagram 
the columns are assumed as partially fixed, the points of contra- 
rlexure being considered as at a distance from the base of the 
column equal to one-third the^ distance from this base to the 
foot of the knee-brace. It is then treated similarly to Case II, 
Chapter III, page 46. The diagram is most readily drawn by 
starting it from the right-hand column. The table on the fol- 
lowing page gives the stresses read from these diagrams. All 
stresses are in pounds. 

Design of Members. — Considering the members EI, JK 
and NO, JK is the longest and carries a load nearly equal to 
that carried by HI; it will require the heaviest section. Its 
loads are — 9900 and + 2500 lbs. Its length is 8.42 ft. 

Trying two 3 in. X z\ in. X \ in.-angles with their long legs 
placed parallel, their least radius of gyration is 0.95 ; this makes 

I = (8.42 x 12) = io6< 

r 0.95 

The allowable fiber stress then is / = 16,000 — (70 X -) = 

8580 lbs. per sq. in. 

The load the two angles will carry is 2.64 X 8580 = 22,600 lbs. 

These angles are therefore ample. 

Member /J. — The length of this piece is 123 ins. and its loads 
are — 10,000 and + 12,800 lbs. Trying two 3 in. X 2 \ in. X 

7 T 22 

T 5 g-in. angles, - = — ^- = 129. Considering the character of the 

loads and the method of attachment this is satisfactory. The load 

it will carry is/ = 16,000 — (70 X -) = 16,000 — (70 X 129) = 

6970 lbs. per sq. in. 

The total load they will carry is 2 X 1.63 X 6970 = 22,700 lbs. 



DESIGN OF A STEEL MILL BUILDING 



129 



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130 GRAPHICS AND STRUCTURAL DESIGN 

Members KL and LM. — These are each 151 ins. long and both 
are liable to be in compression. LM receives the greater load 
which is — 13,600 lbs. Trying two 4 in. X 3 in. X J-in. angles, 

we have - = — *-- = 118, / = 16,000 — (70 X -J = 7740 lbs. 
r 1.28 \ r) 

The total load then is 3.38 X 7740 = 26,160 lbs. 

Members OP and MP. — These are tension pieces and will 
require two 3 in. X 2 J in. X i-in. angles with their long legs 
placed parallel. They will carry in tension (2.64 — 0.44) X 
16,000 = 35,200 lbs. 

These angles are heavy but will be used, being the lightest 
allowed. 

Member MN is a tension piece carrying. + 3900 or + 5270 on 
uniform load basis and one 3 in. X 2§ in. X 4-in. angle will be 
sufficient. Specifications frequently specify that where single 
angles are used as tension pieces and fastened by one leg only, 
such leg shall be able to carry the full load. 

Upon this basis the net area of the 3-in. leg is 0.75 — 0.22 = 
0.53 sq. in. It will carry 0.53 X 16,000 = 8480 lbs. 

Lower chord members, PA y LA } JA and HA. Excepting HA 
these are all purely tension pieces. The maximum force acts 
in HA and is + 63,600 lbs. on the uniform load basis. Try- 
ing two 3 in. X 3 in. X f -in. angles, their net section is found to be 
2 (2.1 1 — 0.33) =3.56 sq. ins. These angles will carry a tensile 
load of 3.56 X 16,000 = 57,000 lbs., and since this may be in- 
creased 25 per cent they will be ample. 

This member is restrained by the knee-brace and its un- 
supported length is about 60 ins. The minimum radius of 

gyration of these angles is 0.01, so that - = = 66. 

r 0.91 

The allowable unit compression is"/ = 16,000 — (70 X 66) = 
11,380 lbs. per sq. in. The total load allowed on the section 
then is 2 X 2.1 1 X 11,380 = 48,000 lbs. The estimated com- 
pression being 2700 lbs. these angles are satisfactory. 

As the lower chord must be made in three pieces for shipment, 
a lighter section might be used for the member PA. 



DESIGN OF A STEEL MILL BUILDING 



131 



Upper Chord. — The upper chord will be made of the section 
shown in Fig. 160. It is first necessary to find the center of 
gravity and then the moment of inertia of 



this section. 




«— -4- 




f— — ' 


Area, sq. ins. Moment. 




_i 


-m 


2 X 2.09 = 4.18 X 0.76 = 3.18 






1 


\ „ „ „ 


9 X 0.313 = 2.82 X 4-5° = 12.65 








2-4x3xV 16 L3 


7.00 15.83 










x — = 2.26 ms. 
7.00 








Plate 


The inertia of the combined section then 




T 


follows : 






Fig. 160. 


Angles, 2 X 1.65 


= 


3-30 




Angles, Ah 2 , 2 X 2.09 X (2.26 - 


- 0.76) 2 = 


9.42 




Plate, M, = a3I2sX9 ' 
12 12 


= 


19.00 





Plate, Ah 2 = 2.81 X (4.50 - 2.26)2 = 14.08 

I = 45.80 

The span of CH is 9.42 ft. When two purlins are used in each 

span of 9.42 ft. the purlin load will be ^^- = 2965 lbs. As- 

2 

suming the purlins centrally located on the above span the 
bending moment is 

M = I fW±) = I X 29 6 S X 9^p = 52 ,37o in. lbs. 



The extreme fiber stress in compression is 
Me w 2.26 



fc 



52,370 x 



45.80 



= 2580 lbs., 



while the extreme fiber stress in tension is 

6-74 



/ _ Me 

ft ~T 



5 2 ,37o x 



7700 lbs. 



45.80 

Combining these fiber stresses we have, 
Total fiber stress in compression side, 

69,800 1U 

- 2580 - -^ = - 12,550 lbs. 

7.0 

Total fiber stress in tension side, + 7700 — = — 2270 lbs. 

7.0 



132 GRAPHICS AND STRUCTURAL DESIGN 

The radius of gyration of this section must now be found 
about the axis through the plate and parallel to the short legs 
of the angles. 

Inertia of two angles referred to axis 2-2, 2 X 3.38 = 6.76 

Ah 2 of the angles, 2 X 2.09 X (1.26 + 0.155) 2 = 8.36 

_ / = i5- 12 

The radius of gyration = r = y —■ = y 5 ' T • = 1.47, 

/ = 16,000 — 70 X - = 16,000 — (70 X — ) = 10,640 lbs. 

r \ 1.47 / 

The maximum allowable stress may be taken 1.25 X 10,640 = 
13,300 lbs. per sq. in. 

The maximum fiber stress in compression being 12,550 lbs. the 
section will do. 

Columns. — The horizontal reactions at the columns will be 
assumed equal so that the horizontal components of Ri and R 2 
will be equal and their sum will be equal to the horizontal com- 
ponent of all the wind forces acting on the structure above the 
points of contraflexure of the columns. The horizontal com- 
ponents of Ri and R 2 are each equal to 5900 lbs. The distance 
from the point of contraflexure to the foot of the knee-brace 
has been assumed 10 ft.; this is two-thirds of the distance from 
the foot of the knee-brace to the base of the column and is taken 
as suggested by Mr. R. Fleming, the writer believing with him 
that this will be ordinarily nearer the truth than either consider- 
ing the column hinged or fixed with the point of contraflexure 
midway between the knee-brace and the column base. Under 
the assumed conditions the bending moment will be a maximum 
on the leeward column and will equal 5900 X 120 = 708,000 
inch pounds. 

The direct load will be the sum of the dead load, the minimum 
snow load and the vertical component of the wind load all act- 
ing on the leeward column. The direct load on the column then 

equals - \- 5000 = 19,600 lbs. We will try four 



DESIGN OF A STEEL MILL BUILDING 1 33 

4 in. X 3 in. X i-in. angles with one 14 in. X iV -in. plate. The 
area of the section is 

4 angles, 4 X 1.69 = 6.76 sq. ins. 

1 plate, & X 14 = 4-37 sq. ins. 

Total 11.13 S( l- * ns - 

Moment of Inertia of Section. — 

4 angles, 4 X 1.4 = S- 6 

Ah 2 of angles, 4 X 1.69 X (7.25 — 0.74) 2 = 286.5 

Plate ,^. ! , A x !i! . 71 . 5 

12 10 12 

Total 363.6 

The extreme fiber stress in the column then due to bending is 

, M X e 708,000 X 7.25 „ 

/ = — - — = J — — - = 14,100 lbs. per sq. in. 

I 363.6 

The column must now be tested to see if this is under the 
allowable stress. The radius of gyration is 



v/f= 


. */363-6 . 
V 11. 13 


= 5-72. 




The direct unit stress is 








19,600 
11. 13 


1760 lbs. 


per sq. 


in. 



The total combined unit stress is 

14,100 + 1760 = 15,860 lbs. per sq. in. 

Assuming the column as having one end fixed, the other sup- 
ported, the allowable fiber stress is 

( 16,000 - 47 X -J X 1.25 = 16,000 - U7 X ^A X 1.25 

= 17,500 lbs. per sq. in. 

The 25 per cent additional stress is allowed in accordance with 
paragraph 90 of the specifications. Some designers allow 50 per 
cent additional here. The 15,860 lbs. per sq. in. is sufficiently 
close to the 17,500 lbs. per sq. in. 

The column should now be examined for its strength referred 
to its other axis. The inertia about the axis parallel to the 



134 GRAPHICS AND STRUCTURAL DESIGN 

column's web is found to 24.4. From this the radius of gyration 

is r = V ■ = 1.48. The unsupported length of this column 
* 11. 13 

being 8 ft., - = — - = 64.8. This value of - being within the 

r 1.48 r 

desired limit the column is satisfactory. 



Wind Bracing 

The reactions at the tops of the end columns, Fig. 161, due 
to wind pressure are 

86 X 20 X 20 , 86 X 19.2 X 20 „ 

1 = 14,100 lbs. 

4 2X3 

The force in the eave strut then is 14,100 lbs. The length of 
this strut is the distance between columns, or 17 ft. 




Fig. 161. 



M 



«"? 



Fig. 162. 



For an eave strut try Fig. 162, composed of four 3 in. X 2 \ in. 
X f-in. angles latticed, the long legs of the angles being turned 
out, 

1 v I2 

- = 17 x — = 145; 

r 1.40 

according to the straight-line formula 

/ = 16,000 — (70 X -] X 1.25 = 7300 lbs. per sq. in. 



DESIGN OF A STEEL MILL BUILDING 135 

In addition to acting as a strut the eave strut replaces a 
purlin so that it is also subjected to bending. The total load on 
a purlin has been estimated as 2965 lbs.; the corresponding load 
on the eave strut will be somewhat less than this depending on 
the amount of overhang of the roof covering. 

The bending moment on this strut then is 

W *L 12 

M = — - — = 2965 X 17 X — - = 75,600 in. lbs. 
8 8 

The distance between the centers of gravity of the strut flanges 
is 21.75 — (2 X 0.66) = 20.43 ms - 

The mean fiber stress in the flanges due to the bending equals 
the bending moment due to the external loading divided by the 
product of the area of one flange (two angles) and the distance 
between the centers of gravity of the flanges. 

, 75,6oo „ 

f = — — — r = 1400 lbs. per sq. in. 

J (2 X 1.32 X 20.43) 

The total resulting fiber stress in the strut then is 

- 14,100 . 1U 

/ = — — + 1400 = 4070 lbs. per sq. in. 

No lighter section can be used as these are the lightest allowed 
by the specifications, for the -value determines the section. 

The transverse brace between the first and second trusses 
will be designed as a simple truss to carry the wind loads at 
its several apexes. ( 

Having determined the apex loads in Fig. 163, the stresses 
in the truss members can be found either by the method of 
coefficients or by making a stress diagram, Fig. 164. The 
diagonals will be assumed as taking only tension. The maxi- 
mum stress is in the end diagonal and is found to be 17,600 
lbs. Trying a 3 in. X 2§ in. X iVin. an gl e > an d assuming one 
rivet hole at a section, the net area is 1.63— 0.27 = 1.36 
sq. ins. 



136 



GRAPHICS AND STRUCTURAL DESIGN 



The fiber stress due to direct loading then is • = 13,000 

1.36 

lbs. per sq. in. This piece is approximately 27 ft. long. Cross 

braces will be used. Their size will be estimated as though but 

one set were in place. These angles will be subjected to bending 

due to their own weight and also due to the attachment being 

made to one leg. The angles weigh 5.6 lbs. per ft. and the 

W - L 

bending moment due to its weight is M = — - — ■ = 5.6 X 27 X 

8 

12 

27 X— = 6124 in. lbs. The bending due to. the tensile force 
8 

acting in the piece is 17,600 X (0.68 — 0.156) = 9222 in. lbs. 




b 



Fig. 163. 



Fig. 164. 



When these act in the same direction their sum is 15,346 
in. lbs. 

The flexural stress resulting is 

M • e 15,346 X 0.68 



f = — M n " , = ~o>o*t~ " — _ T .g Q Uj S> p er S q # m# 

I+ ±*- O . 9 o + ^ 6o ° x ^ 2 

10 X E 300,000,000 

As the allowable unit stress in bracing is 20,000 lbs. per sq. in. 
there is ample margin for initial stress. 

As the stresses in the diagonals of the braces in the top chords 
of the roof trusses and the bracing between the columns are 
both lower than the stress in the above diagonal the 3 in. X 2J 
in. X J-in. angle will be used in all these places. The strut in 
the bracing in the upper chord of the trusses will be stiffened by 
attaching it to the purlin. Here a 3 in. X 3 in. X J-in. angle 
will be used. 



DESIGN OF A STEEL MILL BUILDING 137 

Girts 
The load on the girts is 4 X 17 X 20 = 1360 lbs. 

M = — = 1360 X 17 X -^ = 34,7oo ln - lbs. 

M = I = 34 ? 7oo = 
f e 1.25 X 16,000 

A 5 in. X 2>h m - X xj-in. angle will be used. The struts in the 
transverse truss will be made of two 5 in. X 3! in. X iV m - 
angles, with the long legs parallel. The radius of gyration here 

I . 12 

being 1.46 the value of - is 17 X — 7 = 140. 
r 1.46 

Building of Lighter Design. — In the following problem (Fig. 
165) the stresses due to dead load, snow load and wind load 
have been tabulated and their resultants compared with the 
stresses produced by an equivalent dead load of 40 lbs. per sq. ft. 
of horizontal projection of the roof. 

The building is of somewhat lighter design than the previous 
one. Span of truss, 86 ft. o ins.; rise, \ span. Truss spacing, 
25 ft. o ins. center to center. Figures 167 and 168 are the stress 
diagrams for wind load and dead load. The horizontal wind 
pressure is assumed at 20 lbs. per sq. ft. and the normal pres- 
sure on the roof is taken by Duchemin's formula at 14.9 lbs. per 
sq. ft. The wind is shown as acting upon the left of the truss, 
and the truss members in compression are represented by heavy 
lines. 

The dead-load stress diagram is given in Fig. 168 and has been 
made to serve for both dead load and snow load by changing the 
scale. The snow load has been assumed at 20 lbs. per sq. ft. of 
horizontal projection. The weight of the truss is taken as 

w= £? 



300 + 6 L H 

3 



138 



GRAPHICS AND STRUCTURAL DESIGN 




DESIGN OF A STEEL MILL BUILDING 139 

where 

W = weight of truss per square foot of building. 
L = span of truss, in feet. 

D = distance center to center of trusses, in feet. 
P = load per square foot on truss. 
40 X 86 



W 



300 + (6 X 86) + 4 -^i 
3 



The estimated total weight of one truss is 3 X 25 X 86 = 
6450 lbs. 

The weight of the purlins is assumed as that given by 

V 45 ' 4 

where 

W\ = weight of the purlins per square foot of building, in pounds. 
Pi = load per square foot on purlins, in pounds. 
D = distance center to center of trusses, in feet. 

Jjr V40 X 25 1 „ 

Wi = — - = 3.25 lbs. 

45 4 

The roof covering will be No. 20 galvanized corrugated iron, 
weighing about 2 lbs. per sq. ft. or 2.75 lbs. allowing for laps, 
etc. 

It is commonly specified that roofs of this character shall be 
designed for a uniform dead load of 40 lbs. per sq. ft. of exposed 
surface. This is treated as an equivalent loading and replaces 
the wind, snow and dead loads. 

The stresses will now be determined for the estimated dead, 
snow and wind loadings and their resulting maximum stresses 
compared with stresses determined from the uniform dead load 
of 40 lbs. per sq. ft. 

The apex dead load on the upper chord of the truss at 9 lbs. 
per sq. ft. is 

86 X 25 X 9 , „ 

a = i6iolbs. 

12 



140 



GRAPHICS AND STRUCTURAL DESIGN 



The apex snow load at 20 lbs. per sq. ft. of horizontal projection 

ls 20 X 25 X 86 Q „ 
a = 3580 lbs. 



12 




4-L8 3J£x2^x!£ 
„ S.L;1.L. 



H i i 



Fig. 169. 



The purlins are trussed. 

The apex normal wind load at 14.9 lbs. per sq. ft. of roof surface, 
the distance between the panel points along the upper member 
of the truss being 8.05 ft., is 8.05 X 25 X 14-9 = 3°°° lbs - 



DESIGN OF A STEEL MILL BUILDING 



141 



Member. 


Dead load at 
9 lbs. 


Min. snow load 
at 10 lbs. 


Wind load at 
14.9 lbs. 
(normal) . 


Combined. 


Equivalent 
load at 40 lbs. 
per sq. ft. of ex- 
posed surface. 


CJ 


— 20,250 


— 22,500 


-15,500 


-58,250 


— 101,000 


JK 


— 1,800 


— 2,000 


- 3,48o 


- 7,280 


- 8,970 


LM 


+ 3.240 


+ 3,6oo 


+ 6,600 


+ 13,440 


+ 16,150 


QT 


+ 8,370 


+ 9,300 


+ 13,200 


+30,870 


+ 41,900 


JA 


+ 18,450 


+ 20,500 


+ 22,200 


+61,150 


+ 92,000 


AT 


+ 9.9°° 


+ 11,000 


+ 6,350 


+ 27,250 


+ 49,400 


TV 


— 20,250 


— 22,500 


— 17,100 


-59,850 


— 101,000 


MN 


- 4,320 


— 4,800 


- 8,300 


— 17,420 


— 21,500 



The above figures indicate the truss members designed upon 
a basis of a uniform load of 40 lbs. per sq. ft. of exposed roof 
surface would be considerably on the safe side. 

The selection of the truss members may be made as was done 
in the preceding problem and will not be detailed again. 



Columns 

When knee braces are omitted the building must be stiffened 
in some other way. This may be done by designing the columns 
to withstand the wind loading in each bay on the side of the 
building, by designing the lateral bracing to carry all wind 
pressures to the ends or by assuming a division of this loading 
between the columns and the bracing. Actually the columns 
and bracing always share such loads, but as the proportion 
carried by each depends upon their relative stiffness the actual 
loading becomes a matter of considerable uncertainty. Just 
what assumptions are best to make will depend upon the length 
and height of the building, the assumed wind pressure, the 
manner of securing the tops of the columns to the trusses and 
the bases of the columns to the foundations. The assumptions 
for the design will also depend upon the judgment of the designer. 

All bracing should be given an initial stress when erected, to 
insure its acting promptly when the loading comes upon it. 

The portion of the column above the crane girder carries the 
roof load and will be liable to buckle about an axis parallel to 



142 GRAPHICS AND STRUCTURAL DESIGN 

the web. The load is applied concentrically to the column. 
There may be bending on the column about this axis from wind 
pressure on the end of the building, and from thrust due to the 
crane stopping and starting on its runway which should also be 
considered. 

Below the crane girder the inside angles of the column must 
transfer the crane load, including impact, to the full section of 

the column. As this transfer will occur in a short distance the - 

r 

value may be neglected, hence the angles on this side of the 

column would require a minimum section of - 1 = 4.13 sq. 

16,000 

ins. The two 6 in. X 3^ in. X iV m - angles used give an area 

of 5.78 sq. ins. and should prove ample. 

The roof load and the side of the building carried by the 
columns being both eccentric to the axes of the columns will 
produce bending on them. 

The resultant of these two loads and the crane load will most 
likely be eccentric to the center of gravity of the columns and 
will, therefore, produce a bending moment upon them. In 
addition to these the transverse thrust from the crane and the 
wind load on the side of the building will produce bending on the 
columns about an axis at right angles to the web. The effect of 
these moments will be materially reduced by the restraint at the 
column bases if the columns are fixed at that point. The maxi- 
mum fiber stress resulting from the combination of direct and 
flexural stresses should not exceed that permitted upon the 
column when the maximum allowable stress has been reduced 
by a suitable column formula. Considering that the stresses are 
the resultant of wind load, crane load and dead load this reduced 
value may be increased 25 per cent. The column section below 
the crane girder must also be designed for a possible buckling 
about an axis parallel to the web, and for bending due to the 
crane thrust and wind load on the end of the building. A very 
common practice is to place a channel at right angles to the 



DESIGN OF A STEEL MILL BUILDING 



J 43 



column axis, riveted to the inside flange angles. The channel 
carries the crane load directly from the runway girder. 

The reactions at the tops of the columns are assumed as carried 
to the ends of the building by the lateral trusses in the lower 






Fig. 170. 

LOWER CHORD BRACING 









































END VIEW 

Fig. 170a. 



Crane Girder — * 
Strut 



^ 



,£2 



^ 



^ 



SIDE ELEVATION 

Fig. 171. 



chords of the roof trusses. As members of the lower chords of 
the roof trusses form parts of the lateral bracing such members 
should be examined to see that they will carry these bracing 
stresses in addition to those already on them from the truss load. 
The arrangement of the bracing is shown in Figs. 170 to 171. 



CHAPTER XI 
DESIGN OF A RAILWAY GIRDER 

Specification. — Design the girders for a 68-foot span deck- 
plate girder bridge to carry a single track for Cooper's E-60 
loading, given in the table on page 72. The working fiber stress 
is to be 16,000 lbs. per sq. in. in tension and this is to be properly 
reduced by a straight-line formula for compression members. 
The allowable shearing fiber stress on the rivets will be 12,000 lbs. 
per sq. in., while 24,000 lbs. per sq. in. will be allowed in bearing. 
All rivets will be f in. in diameter and rivet holes will be reamed 
to xf in. in diameter. 

The girders will be placed 7 ft. o ins. center to center. The 
unit shearing fiber stress in the web plate shall not exceed 10,000 
lbs. per sq. in. 

The dead load of the bridge, two girders, shall be assumed of 
the following weights in pounds per foot of span: 

Track 450 

Steel, girders, bracing, etc 1040 

Total 1490 

Allowable pressure between base plate and pier shall be 400 lbs. 
per sq. in. 

Bending Moments. — The dead-load bending on one girder 

WL . ... 

equals at the center, at which point it is a maximum. 

8 

M = E^ = I42? x68x68x H = SjI 6 7 ,320iii.lbs. 

o 2 o 

The maximum live-load bending moment will occur when the 
greatest load is on the girder, and the heavier loads near the 
center. The bending will then be a maximum with the loads so 

144 



DESIGN OF A RAILWAY GIRDER 145 

placed on the girder that the center of the span bisects the dis- 
tance between the center of gravity of the loads on the span and 
the adjacent wheel. The bending will be a maximum under a 
wheel and, there being a wheel on each side of the center of 
gravity, it is necessary to consider each of the two wheels in turn 
as the adjacent wheel to see under which the bending moment is 
the greater. Care should also be taken to see if any of the loads 
in the assumed loading pass from the girder or any additional ones 
come on it, as the position will not then give a maximum moment. 



J2.93' ,,!___ 

u25 — 



jol 34'<£- — - 



Fig. 172. 

With the 68-foot span under consideration to have heavy wheel 
concentrations near the center when the bridge is fully loaded, it 
is necessary to have the drivers of the second locomotive in the 
center of the span. Wheel 13 at the center of the span would 
bring wheels 8 to 18 inclusive on the span. The center of grav- 
ity of this group must now be found, and the table of moments 
on page 72 will be used for this purpose. The moment of 
wheels 8 to 18 about wheel 18 from the table is 7,525,500 ft. lbs. 
and this divided by the sum of these loads, 252,000 lbs., gives 
29.86 ft. as the distance of the center of gravity from wheel 18. 
Now placing these loads on the span in accordance with the pre- 
viously stated conditions for maximum bending we have the 
loading shown in Fig. 172. 

To determine the bending moment we must first find the left 
reaction. Take the moment of the loads 8 to 18 about load 18 
and add to it the moment of the sum of the loads on the girder 
by the distance the right pier is to the right of load 18. The sum 
of these moments divided by the span will give the left reaction. 

R = [7>5 2 5,5°° + (252,000 X 4-07)] -5- 68 = 125,750 lbs. 



or 



146 GRAPHICS AND STRUCTURAL DESIGN 

The bending moment under load 13 is then equal to the 
left reaction multiplied by the distance from the left pier to 
wheel 13 less the moment of the loads on the girder to the 
left of load 13. These moments also are taken from the 
moment table. 

M = (125,750 X 33.93) - 1,831,500 = 2,435,197 ft. lbs. 

M = 2,435,200 X 12 = 29,222,400 in. lbs. 

The moment must now be found under wheel 14 when the 
center of the span bisects the distance between the center of 
gravity of the loads from 8 to 18 and wheel 14. Wheel 18 will 
then be 6.57 ft. from the right pier and wheel 8 will be 0.43 ft. 
from the left pier. The moment calculated as in the preceding 
case gives the bending under wheel 14 as 28,429,000 in. lbs., 
which is less than that found under wheel 13 in the preceding 
case. 

To the maximum live-load bending must be added the dead- 
load bending and the bending due to impact. The customary 
allowance for the latter varies with the designer but will here be 
taken as that given by the formula 

7 _5[_as2_y 

\L + 300/ 

I = impact to be added to the live-load stress. 
■S = calculated maximum live-load stress. 
L = length of loaded distance in feet which produces the 
maximum stress in the member. 

The percentage increase in the live-load stress to cover impact 
in this problem is 

—3 = 81.5 per cent. 

68 + 300 D F 

The maximum impact allowance then is 29,222,400 X 0.815 = 
23,816,256 in. lbs. 



DESIGN OF A RAILWAY GIRDER 147 

The total maximum bending moment becomes: 

Dead-load bending 5,167,320 in. lbs. 

Live-load bending 29,222,400 in. lbs. 

Impact allowance 23,816,260 in. lbs. 

58,205,980 in. lbs. 

In designing a girder it is usually only necessary to determine 
the maximum bending moment, so the above method is all that 
is required. If it is desired to determine the bending moments 
for a number of points along the span this can readily be done 
by the graphical method explained on page 47; this method 
has been used to determine the maximum moment and is shown 
in Fig. 173. This diagram gave the maximum moment as oc- 
curring under load 13 when at the center of the span, and the 
amount of the bending moment was estimated at 29,400,000 
in. lbs., this moment differing from that calculated by less than 
1 per cent. The calculation gave wheel 13 as 0.07 ft. to the left 
of the center of the span which is in very close accord with the 
diagram. 

Maximum End Shear. — A diagram of maximum shears will 
be drawn (see Fig. 175 and explanation on page 49). The end 
shear will also be calculated. The maximum shear on any 
section will generally occur when the span to the right of that 
section is fully loaded. In the case of locomotive wheel loads, 
owing to the considerable difference in weight between the pilot 
wheel 1 and the first driver 2, the maximum shear may occur with 
wheel 2 at the section rather than with wheel 1 there. In this 
case placing wheel 2 over the left pier will bring wheel 13 on the 
span 2 feet from the right pier. The reaction then is 

R = 10,392,000 + (318,000 - 15,000) x 2 = l6l735lbs 

The diagram of maximum shears checks this value of the end 
shear and shows that the maximum shear occurs with wheel 2 at 
the pier instead of with wheel 1 there. 



148 



GRAPHICS AND STRUCTURAL DESIGN 



DIAGRAMS FOR DECK PLATE GIRDER 
Span 68 ft. Loading Cooper's E 60 




Fig. 175. 



Wind Bracing 



Lengths of Flange Plates 

Fig. 176. 



Fig. 177. 



Fig. 178 




Fig. 181. 



Fig. 180. 



DESIGN OF A RAILWAY GIRDER 149 

Dead-load Shears. — The weight per foot of one girder was 
previously estimated at 745 lbs. The dead-load end shear, being 
one-half the weight of the girder, is equal to 745 X - 6 2 8 - = 25,330 
lbs. The dead-load shear varies uniformly across the girder, 
changing from this value at the end of the span to zero at the 

middle; it will, therefore, be D, °° = 12,665 lbs. at the quarter 

point. 

The impact shear will be given by the formula that was used 

for impact bending, / = S[ T — ) . This gives the impact as 

\L + 300/ 

81.5 per cent of the live-load shear when the girder is fully loaded, 

and the impact shear becomes 0.815 X 161,735 = i 3 i j8io lbs. 

Combining the several end shears we have: 

End shear, dead load 2 5>33Q lbs. 

End shear, live load 161,735 lbs. 

End shear, impact 131,810 lbs. 

318,875 lbs. 

Allowing a unit shearing stress of 10,000 lbs. per sq. in. requires 
a web area of 31.9 sq. ins. A web plate 80 ins. deep and y 7 g in. 
thick has an area of 35.0 sq. ins. and is ample.* 

Flange Area and Selection of Sections. — It is common 
practice to assume the effective girder depth, that is, the distance 
between the centers of gravity of the girder flanges, as approxi- 
mately the distance back to back of the flange angles. After the 
flange sections are chosen their centers of gravity are determined 
and this distance compared with the distance first assumed; if 
the error is small no correction is made, otherwise the distance 
is taken as the calculated distance between the centers of gravity 
of the flanges and new flanges are determined and a new selection 
of sections is made. 

The girder has been assumed as having an 80-in. web plate and 
the angles will be set out J in. on each side from the edge of the 
plate to avoid the possibility of the plate extending beyond the 
angles. The effective depth will be taken as 80.5 ins. The 

* See Specifications, § 100 to § 154. 



i5° 



GRAPHICS AND STRUCTURAL DESIGN 



allowable working fiber stress being 16,000 lbs. per sq. in. the 
net flange area is found by dividing the maximum bending 
moment in inch pounds by the product of the allowable fiber 
stress and the girder depth. 

Net flange area = — — — ^- — = 45.2 sq. ins. 
16,000 X 80.5 ° H 

This area will be made up of the following sections: 

One-eighth web area = | X 35.0 = 4.4 sq. ins. 

Two 6 X 6 X f-in. angles, less 4 rivet holes 13 . 9 sq. ins. 

Three 14 X f-in. plates, less 2 rivet holes 27 . o sq. ins. 

Total area 45 . 3 sq. ins. 

This is sufficiently near the 45.2 sq. ins. required. The rivet 
holes have been assumed as 1 in. in diameter. As the calculated 
distance between the centers of gravity of the flanges exceeds 
that assumed by 0.20 in. the section assumed will be slightly 
larger than required, about \ per cent. This is an unimportant 
difference and will be neglected. 

The live -load shears will be taken from the diagram of maxi- 
mum shears, Fig. 175, and to these shears will be added the dead- 
load and impact shears. 





Shears in pounds. 




End. 


\ span. 


§ span. 


Dead load 


25.330 
162,000 
131,800 

319.130 


12,665 

99,000 

82,500 

194,165 


O 


Live load 


46,000 


Impact 

Total shear 


40,400 
86,400 







At the quarter point, when wheel 2 is at this section, the 
loading extends on the girder 59 ft. from the right pier and 
according to the formula the percentage of live-load shear to be 
3°o 



allowed for impact is 



83.5 percent. The impact shear 



59 + 3°° 

is 99,000 X 0.835 = 82,500 lbs. In the same way the allowance 
for impact at the center of the span is found to be 40,400 lbs. 



DESIGN OF A RAILWAY GIRDER 151 

Stiffening Angles. — The end stiffeners act as columns, but as 
the load is shared by the web and as their length is short they are 
generally estimated by using the allowable fiber stress but not 
reducing it by a column formula. The area required in the end 
stiffeners then is 



Max. end shear _ 319,130 _ 



16,000 16,000 



9.93 sq. ins. 



Four 5 in.X 3 J in.X r£- m - angles, having a combined area of 4 X 
5.38 = 21.52 sq. ins., will be used. The intermediate stiffeners 
are not usually calculated. Angles of the same dimensions but 
of lighter section are used. The intermediate stiffeners will be 
made two 5 in. X 2>i m - X f -m - angles. 

The four end stiffeners being over the end bearing plate, the 
rivets will be assumed as transferring the end shear from the web 
to the four angles ; that is, the load will be assumed as distributed 
over the entire number of rivets in the two pairs of stiffeners. 
The rivets will be f in. in diameter and will be in double shear 
and will bear on the y^-in. web plate. Allowing a shearing fiber 
stress of 12,000 lbs. per sq. in. and a bearing fiber stress of 
24,000 lbs. per sq. in. the rivet value in double shear is 14,430 lbs. 
while the value in bearing is 9190 lbs. The number of rivets 

required in the two pairs of end stiffeners is ^ "' ^ =35- 

Lengths of Flange Plates. — The flange-plate lengths can be 
approximated as in Fig. 176. Lay off a line representing the 
span of the girder to scale and at its center erect a perpendicular 
representing to scale the calculated net flange area, in this case 
45.3 sq. ins. Now upon the span as a base and with the 45.3 as 
an altitude draw a parabola. The parabola construction is shown 
in the light lines. Starting at the base line lay off in succession 
distances representing one-eighth the web area, 4.4 sq. ins. ; the 
net area of the two flange angles, 13.9 sq. ins.; and the net area 
of the three flange plates, 9 sq. ins. each. The distance measured 
along the lower line of any section between the sides of the 



152 GRAPHICS AND STRUCTURAL DESIGN 

parabola gives the theoretical length on the assumption that 
the bending moment across the girder varies as the ordinates in 
the parabola. This is not exact, although approximately true. 
Were greater accuracy desired the bending moments could be 
estimated for intervals along the girder and the .corresponding 
moment diagram drawn, which could then be used instead of 
the parabola. A close approximation for a locomotive and train 
load may be made by drawing a rectangle for the center of the 
diagram, Fig. 176, whose width is one- tenth the span and whose 
ordinate represents the net flange area. On each side of this 
rectangle a half parabola is then drawn. 

'It is customary to take lengths of plates slightly exceeding 
the lengths determined in the way just described. The follow- 
ing are the scaled lengths and the lengths used. 



Outside plate, top and bottom. 
Middle plate, top and bottom. 

Inside plate, top 

Inside plate, bottom 




Flange Rivets. — First consider the rivets in the vertical legs 
of the flange angles securing these angles to the web plate. 
The rivets transfer the change in horizontal shear from the 
flanges to the web plate. The flange forces acting at intervals 
along the girder can be determined and the change in this force 
between adjacent sections used to estimate the number of rivets 
required. The change in flange force divided by the rivet value 
will give the number of rivets between the two sections. Be- 
sides this change in horizontal shear the rivets also transfer the 
vertical loads to the web through the upper flange rivets. The 
maximum wheel concentrations are usually considered as dis- 
tributed over three ties or approximately 36 ins. along the flange. 

These two shears can be combined in the following manner: 
Find the change in flange force per inch of span by dividing 
the difference between the flange forces acting at two adjacent 



DESIGN OF A RAILWAY GIRDER 153 

sections by the distance between these sections; this will give 
the average change of flange force per inch of span between the 
sections chosen. The vertical shear transferred from the angles 
to the web plate per inch of span will be the maximum wheel 
concentration divided by 36 ins. The square root of the sum of 
the squares of these two quantities will give the resultant shear 
per inch of span within the given section and the rivet value 
divided by this resultant shear per inch will give the rivet spacing 
in this section. 

The rivet value referred to is the lower of the two values in 
bearing or shear. 

The following is the commonly used and less tedious method 
of determining the flange riveting. The rate of change in flange 
force per inch at a section is given by 

'4 

where 

V = maximum vertical shear in pounds at the section. 
h = the distance between rivet rows in top and bottom 
flanges, measured in inches. Where there are 
double rows of rivets in the vertical legs of the 
flange angles h is the average distance. 

Where the web is assumed as resisting bending the value / 

should be reduced as the rivets are not required to provide 

strength to secure that part of the flange to the web which is 

already an integral part of it, that is, the one-eighth web area. 

A = total net area of flange, and 

a = one-eighth web area, then the reduced force becomes 



h A 

As before the vertical shear per inch transferred by the rivets 
due to the maximum wheel concentration is 
, Max. wheel load 
3 6 



154 GRAPHICS AND STRUCTURAL DESIGN 

and the resultant shear per inch of span at the section under 
consideration is 

/• = V/ X » + / 2 2 . 

The rivet spacing then is 

Rivet value 
P = 7 

The rivet spacing is generally made the same in the two 
flanges. 

The rivet spacing will now be estimated for the ends of the 
girder. 

The average distance h in this case is 80.5 — (3 X 2.25) = 

73-75- 

/l= Z x ^ = 3IM3o x (4-4 + i3.Q + Q)-4.4 = 6 lbs< 
h' A 73.75 4.4 + 13.9 + 9 

30,000 



h = Q - ± ^~ = 835 lbs. f z = V3630 2 + 8 3 5 2 = 3730 lbs. 

The rivet value in bearing for J-in. rivets at 24,000 lbs. per 
sq. in. is 9190 and the rivet spacing then is p = fyf-g- = 2.46 ins. 
Use 2^-in. spacing. 

Rivet spacing a distance of one-quarter the span from the piers, 

, V ..A — a 194,000 X 31.9 „ 

/1 = — X — - — = - za:2 7—^ = 2310 lbs. 

J h A 73.75 X 36.3 6 

As found before, 

f 2 = 835 lbs. 



fz = V2310 2 + 835 2 = 2460 lbs. 
P = UU = 3-73 ins. 
Rivet spacing at the center, 

= Vx(A-a) = 86,400 X 40.9 = lbs _ 

J hxA 73.75 X 45-3 

ft = V10S5 2 + 835 2 = 1350 H>s. 
P =AH% = 6.8 ins. 



DESIGN OF A RAILWAY GIRDER 155 

The rivet spacing may be calculated for the several panels by the 
method just shown or it will be given with sufficient accuracy 
by laying off, at right angles to a line representing the span, 
ordinates showing to scale the rivet spacing at the ends, the 
quarter points and the middle of the girder and then connecting 
the adjacent points with lines. The spacing at any section may 
be found by scaling the distance from the axis to the line just 
drawn at the particular section. 

Riveting of the Flange Plates. — From what has been stated 
concerning riveting the legs of the flange angles to the web plate 
it will be evident that the rivets connecting the flange plates to 
the angles are called upon to do much less. A common practice 
is to put in rivets securing the plates to the angles and have them 
stagger the rivets in the vertical legs. This prevents the inter- 
ference of the rivets and furnishes more than sufficient rivets. 
Where there is no liability of interference this practice need not 
be followed. The number of rivets required to secure a plate in 
place and develop its strength can be estimated by multiplying 
the net area of the plate by its working fiber stress and dividing 
this by the proper rivet value. In this girder the plates are 
14 X f in. and the net area is 9 sq. ins.; at 16,000 lbs. per sq. in. 
the force in the plate is 144,000 lbs. and the rivets being f in. 
in diameter and in single shear their value is 7215 lbs., which is 
less than the bearing value and must, therefore, be used. The 
number of J-in. rivets in single shear to transmit 144,000 lbs. is 

144,000 

-^ = 20. 

7215 

The middle plate being 46 ft. long and the top plate 33 ft. long, 
the middle plate extends 6.5 ft. beyond the top plate at each end 
and the full strength of this middle plate must be developed in 
6.5 ft. 

The rivet spacing, considering that the rivets are in a double 
row, must not exceed 



6.5 X 12 Q . 

— J =7-8 ms. 

10 



156 GRAPHICS AND STRUCTURAL DESIGN 

The spacing ordinarily does not exceed 6 ins. and in this class 
of work is frequently specified not to exceed 4 ins. The rivet 
spacing is also commonly made less at the ends of the plates and 
opposite open holes left for field rivets. 

Bracing. — The lateral horizontal bracing will be designed in 
accordance with paragraph 144 of the specifications. 

Since only one lateral girder will be used the load per foot of 

girder will be 1000 lbs. The maximum stress will occur in JK 

when the bridge to the right of KL is loaded. The shear in JK 

then is 

(68 — 5.23) X 1000 X 62.77 „ 

° °' n LL = 20.000 lbs. 

68 X 2 y ' 

an, • 7r . 29,000 X 8.74 , „ 

The stress m JK is — '— = 36,200 lbs. 

7.00 

Trying one 6 in. X 4 in. X §-in. angle, the direct stress is 

36,200 / U_ 

i2 — L = 7600 lbs. per sq. in. 

475 
Riveting by the longer leg the allowable fiber stress is 

/ A l\ I , 70 X 96A 

1.25 (l6,000 — 70 "J = 1.25 [l6,000 — J — I 

= 12,700 lbs. per sq. in. 
The stress due to the bending is 

. Me 36,200 X 0.74 X 0.99 _ 26,500 

7 _ W ~ 6 36,200 X 96 2 5-2 

10 E ' 10 X 30,000,000 
= 5100 lbs. per sq. in. 

The combined fiber stress is 

7600 + 5100 = 12,700 lbs. per sq. in. 

which agrees with the allowable. 

This piece will require at least seven f-in. diameter rivets, 
these being in single shear, with a rivet value of 7220 lbs., and 



DESIGN OF A RAILWAY GIRDER 157 

being field driven requiring an allowance of 25 per cent. The 
calculation being as follows: 

^— l — = 6.3, say 7 rivets. 

7220 

End Frames. — The stress in BJ will be assumed as one-half 

the reaction or = 17,000 lbs. This piece is about 

2X2 " F 

75 ins. long. 

The diagram of the end-bracing is given in Fig. 178, the force 
triangle is drawn in Fig. 181, and this gives the stress in the 
diagonal at 23,600 lbs. The diagonals will be fastened together 
at their point of intersection. 

These lengths and forces being less than those for JK the 
minimum weight 6 in. X 4-in. angles will be more than ample. 

Intermediate Cross-Frame. — The estimated load on the upper 

* * * 4.1. t • 68 X 3 X 800 „ T ^ 

strut of the cross-frame is - — = i2,c;oo lbs. Its 

length is approximately 75 ins. The least radius of gyration 

should be -^- = 0.63. This suggests a 3I in. X 3 J in. X f-in. 
120 

angle, being the smallest allowed. This angle will carry 
1.25 (16,000 — • — 7~j X 2.48 = 26,100 lbs. These angles 

should be ample for the intermediate cross-frames. 

Riveting. — The number of rivets between the girder and the 
plate connecting the horizontal bracing members may be de- 
termined as shown in Fig. 179. 

Lay off on a line parallel to PQ a distance representing the 
number of rivets in PQ, and similarly for RS; as one of these is 
in compression, the other in tension, their components will be in 
the same direction along the main girder flange, and hence the 
number of rivets between flange and plate can be determined by 
projecting these lengths upon the horizontal line which here 
scales 10 rivets. 

Web Splice. — When the web is assumed as only resisting shear 
the splice is calculated to provide for the maximum shear at the 



158 GRAPHICS AND STRUCTURAL DESIGN 

spliced section. When, however, the web is assumed as resisting 
bending, that is, one-eighth of the web area is taken as acting 
at the flanges, then the splice plates and rivets must be designed 
to resist a corresponding bending moment. The depth of the 
splice plate will be the distance between the edges of the verti- 
cal legs of the flange angles. In this case, allowing J in. for 
clearance, the depth of the splice plate will be 80.5 — (12 + 2) = 
68 ins. 

The fiber stress at the center of gravity of the flanges was 
assumed as 16,000 lbs. per sq. in.* The distance from the neutral 
axis of the girder to the center of gravity of a flange was assumed 

as - — — = 40.25 ins. The fiber stress at the top of the splice 

2 

plate will be (16,000 X -2 8 -) -f- 40.25 = 13,500 lbs. per sq. in. 

The bending moment resisted by the web is X depth 

8 

of girder X fiber stress = M = 3 f- X 80.5 X 16,000 = 5,635,000 

in. lbs. The area of the splice plate required to resist this 

moment is determined by a calculation similar to the one made 

for the web -plate. This area is 

. _ 8 X bending moment at section, in. lbs. 

depth of splice plate, ins. X extreme fiber stress, lbs. per sq. in. 

. 8 X 5,635,000 

A = , n = 49.0 sq. ins. 

68 X 13,500 ^ 4 

The thickness of each of the two splice plates then is 

t = -42- = 0.36 in. 
2 X 68 6 

If the rivet farthest from the neutral axis develops its full 

rivet value the stress upon any other rivet will be proportional 

to its distance from the neutral axis and the moment due to any 

rivet being also proportional to this distance the total resisting 

moment of a single row of rivets similarly arranged about the 

neutral axis is 2 X rivet value ^ 9 

yi 

* Note. — This is for the center of the span, elsewhere the fiber stress should 
be estimated. 



DESIGN OF A RAILWAY GIRDER 



159 



Here 

M = resisting moment in inch pounds of one vertical row of 

rivets. 
R = rivet value, in pounds. (The lower of the values in 

bearing or shear should be used.) 
y = distance of the rivet from the neutral axis, in inches. 
yi = distance from the neutral axis to rivet in splice plate 
farthest from the neutral axis. 

Assuming the following distances of the rivets in one row above 
the neutral axis, the following values are calculated for y 2 in 
Fig. 182. 




o o 
o o 
o o 
o o 

o o 
o o 
o o 
o o 
o o 
o o 
e-e 



t-e-e- 





Fig. 182. 



Fig. 183. 



Number of 






Number of 






rivet. 


y 


y 2 


rivet. 


y 


y- 


I 








7 


18 


324 


2 


3 


9 


8 


22.5 


506.25 


3 


6 


36 


9 


26.5 


702.25 


4 


9 


81 


10 


29-5 


870.25 


5 


12 


144 


11 


3 2 -5 


1056.25 


6 


IS 


225 






3954 .00 



The rivet value of a |-in. rivet bearing in a ^-in. plate is 

9190 lbs., hence the moment of one row of rivets spaced as 

assumed is 

,, 2 X 9190 w . „ 

M = z - — X 3954 = 2,240,000 m. lbs. 

3 2 -5 



160 GRAPHICS AND STRUCTURAL DESIGN 

The number of rows required on this basis is 

S,6^s,ooo , 

&-^ = 2\ rows. 

2,240,000 

This would take three rows or a rearrangement of the rivet 
spacing to increase the number of rivets in the vertical row and 
at the same time the moment of the rivet group. 

There are several other forms of splices, one of which, see Fig. 
183, consists in replacing the one-eighth of the web area by straps 
placed on each side of the flange angles. The area of the strap 
is to one-eighth web area as the square of the distance between 
the centers of gravity of the flanges is to the square of the dis- 
tance from center to center of the straps. If the straps are 9 ins. 
wide the distance center to center of straps will be 59 ins. 

c+ 4.4 X 80.5 2 . 

btrap area = ^-^ — — = 8.2 sq. ins. 

59 2 
The number of rivets on each side of the strap will be 

8.2 X 16,000 

■ = say 15 rivets. 

9190 

The central plates are designed to resist the maximum vertical 
shear at the section. 

Still another method is to splice the web where excess material 
in the flange makes provision for all or part of the bending as- 
sumed as taken by the web and then design the splice for the 
maximum shear and any bending not otherwise provided for 
occurring at the section. 

In the problem in hand the web splice is made 20 ft. from the 
piers, and the shears at this point are dead-load shear 10,300 lbs., 
live-load shear 90,000 lbs., and the impact allowance 76,000 lbs., 
making a total shear of 176,300 lbs. This would require twenty 
f-in. rivets to provide for the maximum shear at this point. 
There is more than sufficient surplus material in the flange (see 
Fig. 176) to care for the bending assumed as coming upon the 
web, so that a splice designed for shear only would prove ample. 



DESIGN OF A RAILWAY GIRDER 



161 




1 62 GRAPHICS AND STRUCTURAL DESIGN 

Flange Splices. — For the ordinary girder spans splices in the 
angles and plates of the flanges are not quite so likely to be 
required as in the web. When such splices are required the same 
general principles may be followed as in the case of the web 
splice. Splice plates may be placed where angles or plates are 
cut or the cut may be made where the flange plate would other- 
wise end and then the splice may be made by continuing the 
flange plate beyond the cut. This affords one of the neatest 
ways of splicing the angles or flange plates. 

Bed Plates. — The area of the bed plate allowing 400 lbs. per 
sq. in. on the masonry will be the reaction divided by 400 or 

^i32 = 798 ^ ins . 

400 

In small girders carrying light loads the connection between 
the girder and the masonry may be made by the use of a couple 
of rolled plates. These should extend but a few inches beyond 
the flange plates as the usual depth of about 1 in. is not stiff 
enough to transmit much load to the pier beyond the edges of 
the flange plate. The bed plate rests upon the masonry while 
the sole plate lies on the bed plate and carries the girder. At 
one end the girder is held firmly in place by both sole and bed 
plates having round holes a little larger than the foundation 
bolts. At the other end the holes in the sole plates are slotted, 
permitting the movement of the sole plate across the bed plate 
due to the expansion and contraction of the girder. The plates 
at the moving end should be planed. 

Where heavy loads are carried upon girders of short span, 
greater depth of bed plates may be required; this type of cast- 
iron bed plate is shown in Fig. 188. 

In the longer spans, exceeding 75 to 80 feet, greater provision 
must be made for expansion and also more care used to distribute 
the load on the masonry. One way of doing this is shown in 
Fig. 189. Here both ends of each girder are carried on pins, 
while one end rests on friction rollers. The pin assists the distri- 
bution of the pressure uniformly over the rollers, if the webs of 



DESIGN OF A RAILWAY GIRDER 



163 



the pedestal are of ample depth. The method of calculation 
would be as follows: Assuming the rollers 4 ins. in diameter the 
allowable load per inch of roller length is L = 600 X d, where 
L = load in pounds and d = diameter of roller in inches. L = 
600 X 4 = 2400 lbs. 

The combined lengths of all the rollers in the nest is 3 9 ' 3 = 

2400 

133 ins. Making the rollers 23.5 ins. long would require— ^~ 

2 3-5 

6 rollers. The bearing area between pin and web is ^ ^ 3 _ 

24,000 




m 



]~ ]- — ^ot<^ &^<yj^^>-f^( o)Co ^f&)-& 



2354- 



Fig. 189. 



13.3 sq. ins. If three i-in. web plates are used the projected area 
is 3 X 1. 00 in. X 6 ins. = 18 sq. ins., and the area will be 
more than ample. 

The web plate is 45 ins. long and the load is to be distributed 
over it. A rough approximation of the depth of the web from 
the bottom of the pin to the lower edge of the web plate may 



be made by considering it a beam section. M = 



IT' 



8 J e 



164 



GRAPHICS AND STRUCTURAL DESIGN 



= / 



b-d 2 



here d = depth of the web in inches, b = combined 



thickness of the webs in inches. 



d 2 



M = 319,130 X ^p = 16,000 X 3.00 X 7- , from which d = 
8 6 

15 ins. In this calculation no account has been taken of the 

angles and plate secured to the lower edge of the web plates, so 

that the 15-in. depth used should prove ample. The rollers are 

kept in place by two flats held by three rods shown at the ends 

and the middle. Angles at the sides secured to the bed plate 

prevent the rollers from moving laterally. The bending on the 

pin is 

^19,130 X 1.25 . ,, 

v ' J a = 133,000 in. lbs. 




Fig. 189a. 

On the basis of an allowable extreme fiber stress in pins of 
22,000 lbs. per sq. in. a pin 6 ins. in diameter will resist a bend- 
ing moment of 466,500 in. lbs., so that in this way the pin is 
exceedingly strong. 

The methods indicated illustrate a couple of the simplest 
methods used for securing girders in place. 

Figure 189a is a very common type made of steel castings, 
the metal ranging from if to 2\ ins. thick. This design permits 
a pin of smaller diameter and a shallower bearing. 



CHAPTER XII 
CRANE FRAMES 

Figure 190 represents a frame for an underbraced jib crane. 
The live load it is to carry is 3 tons (6000 lbs.). To this must be 
added an assumed weight of block, trolley, chain, etc., and this 
will be taken at 500 lbs. The total live load thus becomes 
6500 lbs. The dead load is made up of the weight of the frame 
and hoisting machinery and will be assumed at 2000 lbs. The 
line of action of this dead load will be taken at one-quarter of 
the effective radius from the mast, or 11.75 X 0.25 ~ 3.0. The 
equivalent load at apex 1 will be the load that would produce the 
same moment on the mast, and is 2000 X i 3 q = 600 lbs. 

The chain pull at different points in its length will be estimated 
upon an assumed efficiency of each sheave of 97 per cent, thus 
above the jib the chain pull will be 6500 -f- (2 X 0.97 2 ) = 3460 
lbs., while parallel to the mast it is 6500 -s- (2 X 0.97 3 ) = 3560 
lbs. The pull on the racking chain will be assumed at 1000 lbs. 

The maximum direct stress will occur in the bracing when the 
load is at its maximum radius. To determine the direct stresses 
graphically the apex loads must be found. To calculate the 
equivalent apex load at 1 (Fig. 191), when the load is at its 
maximum radius, take moments about point 3. The equivalent 

apex load at 1 = 6500 X — — = 7650 lbs. 

10 

The upward reaction at point 3 is 7650 — 6500 = 1150 lbs. 

The diagram (Fig'. 192) gives the combined live -and dead-load 
stresses; these can be determined with greater accuracy if done 
separately but this diagram is sufficiently accurate in this instance. 

The jib will have to be considered for bending both when the 
load is at its maximum radius and when the trolley is between 

165 



1 66 



GRAPHICS AND STRUCTURAL DESIGN 



Fig. 190. 




Fig. 193. 



Fig. 194. 



Fig. 195. 



CRANE FRAMES 167 

points i and 2. Here the span is 60 ins. and the distance center 
to center of the trolley wheels is taken as 24 ins. Although 
commonly discussed as a central load the position of maximum 
bending will here be assumed accurately ; that is, when a wheel is 
one-quarter the distance between the wheels from the center of 
the span. Here this distance is \ 4 - = 6 ins. (see Fig. 193). The 
load of 6500 lbs. being carried upon four wheels the load on two 
wheels or an axle is ^-A^ — 3 2 5° lbs. The reaction 

Rl = (325° x 24) + (3250 x 48) _ 39QO lbs _ 

60 
Ri = 6500 — 3900 = 2600 lbs. 
The direct stress in the section of the jib carrying the trolley is 
given by Fig. 194 as 6600 lbs. 

Fiber Stresses. — Crane frames are liable to considerable 
shock. The usual method of caring for this is to take a lower 
working fiber stress than is ordinarily used in structures not sub- 
jected to shock. These fiber stresses run from 10,000 lbs. to 1 2,000 
lbs. per sq. in. for mild steel, which is the material commonly used 
for crane frames. The allowable stress in columns is the above 
properly reduced by a suitable column formula. As it is difficult, 
if not impossible, to stiffen some parts of crane frames — the jib, 
for instance, in the present problem — the allowable stress in 
members acting as beams must be reduced to prevent buckling 
of the compression flange; see curves, page 102. 

Selection of Jib. — The load at the maximum radius creates 
a direct stress in the jib of 13,900 lbs., in tension. The hoisting- 
chain pull is 3460 lbs., while the pull in the racking chain is 
1000 lbs. ; this creates a force of 1000 lbs. acting in the jib when 
the trolley is pulled back and of 2000 lbs. when the trolley is 
drawn forward. The average height of the two racking chains 
from the center of the jib is taken as 8 ins. Calling tension +, 
and compression — , we have : 

Direct stresses in jib. 

Stress due to live and dead loads + 13,900 lbs. 

Stress due to hoisting rope — 3,460 lbs. 

Stress due to racking rope — 2,000 lbs. 



1 68 GRAPHICS AND STRUCTURAL DESIGN 

The greatest stress then is the combination of the stress due to 
live and dead loads with the stress due to hoisting. The racking 
force will not be considered as it gives a lower stress than the 
above combination. The maximum stress is 13,900 — 3460 = 
10,440 lbs. in tension. 

The bending moment is (6500 X 21) — (3460 X 14) = 
88,060 in. lbs. The stress must now be considered when the 
trolley is between points 1 and 2. -The maximum bending = 
Ri X 24 = 2600 X 24 = 62,400 in. lbs., due to the hook load. 
To this must be added the bending due to the hoisting- and 
racking-rope pulls. The former is M 2 = 3460 X 14 X 0.5 = 
24,200 in. lbs. That due to the latter is M% = 2000 X 8 X 0.5 
= 8000 in. lbs. The total bending is 62,400 + 24,200 + 8000 = 
94,600 in. lbs. The direct stress is 6600—3460 — 2000 = 1140 lbs. 

The influence of span to flange width must now be considered. 
Trying 8-in. channels at \\\ lbs. per foot the flange width is 
2.26 ins., and the span being 60 ins. it follows that 
Span = _6o_ = ^ 
Flange width 2.26 

On the cantilevered section, 

Span 21 

■ — ; = = 9.3. 

Flange width 2.26 

From the curve, page 102, the allowable stress should not exceed 
85 per cent of the maximum desired, so that allowing a maxi- 
mum working fiber stress of 10,500 lbs. per sq. in. it reduces to 
10,500 X 0.85 = 8900 lbs. per sq. in. 
The section modulus for an 8-in. channel at n J lbs. is 8.1; 

hence, since M = f- and the bending has been calculated for 
two channels, we have 



- Me 94,600 ,, 
f = — r = ■ ' ■ = 5840 lbs. 
J I 2X8.1 M 

>. per square inch = 

2 X 3-3^ 
8-in. channels at nj lbs. will be satisfactory. 



The direct stress, per square inch = =170 lbs. The 

2 X 3-35 



CRANE FRAMES 1 69 

Member AH. — The length of this piece is 7 ft. and the 
force acting in it is 16,100 lbs. This member is in compression 

and as it cannot be braced and its - value must be limited to 140 

r 

the least radius of gyration will have to equal or exceed y^o = 

0.60. Seven-inch channels will be required and their radius of 

gyration is 0.59. According to the abridged form of Ritter's 

formula for soft or mild steel 



/1 = L = io ^°° = io ^°° = -„ olbs 

nv t , i 9; 6oo 2.96 



x- 1 + 



10,000 \r/ 10,000 

and the total allowable load on two 7-in. channels at 9! lbs. per 
foot or 2.85 sq. ins. in the section is 3550 X 2 X 2.85 = 20,300 
lbs., which is satisfactory. 

Member AF. — This is a compression piece carrying a load 
of 17,700 lbs.; its length is approximately 13 ft., but, as it can be 
braced across, its greatest length need not exceed the length of 
the piece HA , and as the force acting in it is about the same as 
in the preceding piece it will be made of the same section, 7-in. 
channels at 9! lbs. per ft. 

Member FG. — This is also a compression piece; its length 
is 60 ins. and the force acting in it is 10,500 lbs. Assuming as 

before that the - value shall not exceed 140, r must not be less 
r 

than t 6 3°q- = 0.43. Trying 5-in. channels at 6| lbs., which have 

r = o.^o, we find- = ■ = 120. Substituting, as before, in the 

r 0.50 

column formula gives 

_ xo,5oo _ 10,500 _ ^ 

1 + _i_ x £ 2 2-44 
10,000 \r/ 

The total allowable force on the piece is 4300 X 2 X 1.95 = 
16,700 lbs., and these channels will be used. 

Mast. — For most positions of the trolley the mast will be in 
tension. The position of maximum compression will require the 



170 



GRAPHICS AND STRUCTURAL DESIGN 



trolley to be placed as close to the mast as possible. The sever- 
est stress on the mast will be due to the bending. 

The horizontal reaction at the upper pin can be found by taking 
moments about the lower pin. 

R X 13.75 = ( 6 5°° X n.75) + (2000 X 3.0) = 6000 lbs. 

The bending at the upper portion of the mast equals 6000 X 
22 = 132,000 in. lbs. The mast also resists bending due to the 
5 -in. channels which fall below the line of intersection of the other 

I 



Ri« 6230- 

i 




132000" ^ 




_LU 

Fig. 196 Fig. 197. Fig. 198. 

pieces at apex 3. The reaction on the mast due to this force in 
FG, the horizontal component of which is 6800 lbs., is 



Ri = 



6800 X 11 



12 



= 6230 lbs. 



and the bending is 

M = 6230 X 12 = 74,760 in. lbs. 
This bending is in the opposite sense to that due to the force on 
the pin. 

The bending due to the hoisting-rope pull must now be deter- 
mined. The horizontal reaction at apex 3 due to the rope pull on 
the drum is found by taking moments about apex 5, from which 

= 2560^ _ lbs _ 
12x12 /o 



CRANE FRAMES 171 

The maximum bending due to the rope pull then is 
M = 173 X 90 = 15,570 in. lbs. 
Fig. 198 shows the maximum bending on the mast to be 

132,000 in. lbs. 

„. ,, ,1 I M 132,000 

Since M=f-, - = -7- = - s2 - J = 12.5. 

e e f 10,500 

This is for two channels, or 6.25 for one, and would require 7-in. 
channels were there no direct stress. Try 8-in. channels, the 
area of two such channels being 2 X 3.35 = 6.7 sq. ins. The 
direct stress previously found was 9100 lbs. and the unit fiber 

stress is —, — = 1350 lbs. per sq. in. The fiber stress allowed for 
6.7 

bending then is 10,500 — 1300 = 9150 lbs. per sq. in. It follows 

that 1 M 132,000 

- = — = -^ — ■ = 14.4. 
e f 9150 

This will take two 8-in. channels at nj lbs. whose section 
modulus is 2 X 8.1 = 16.2. 

The resulting moment at any section is the intercept between 
the heavy fines in Fig. 198. The small moment at 5 has been 
omitted as it would not alter the maximum at 3. 

Design of Frame for Top-braced Jib Crane 

The capacity of the crane is to be 5 tons (10,000), at a maxi- 
mum radius of 20 ft. The weight of the trolley, hook and chain 
will be assumed at 600 lbs. The dead load of the frame and 
machinery will be taken as 3000 lbs., and its center of gravity, or 
line of action, will be considered as one-fourth of the crane radius 
from the mast, or -^- = 5 ft. The load which, acting at apex 1, 
would produce the same moment on the mast as the frame 
weight is ^°-^ = 750 lbs. 

The maximum fiber stress in tension is to be 12,000 lbs. per 
sq. in. In compression the maximum fiber stress will be 12,000 
lbs. per sq. in., properly reduced by a column formula. The 

value of - will be limited to 140. 
r 



172 



GRAPHICS AND STRUCTURAL DESIGN 




The skeleton of the frame is shown in Fig. 199, while the 
stress diagram is represented by Fig. 201. The total equivalent 
load at apex 1 is the sum of the live load, 10,000 lbs., the weight 
of the trolley, hook and chain, 600 lbs., and the equivalent frame 



CRANE FRAMES 173 

load, 750 lbs., making a total of 11,350 lbs. The horizontal re- 
actions at points 4 and 6 can then be found by taking moments 
about apex 6. 

„ n,35o X 20 011. 

R = — — = 12,800 lbs. 

17-75 

These several external forces can now be located at the proper 
points and the stress diagram drawn in accordance with the 
principles given in Chapter II. 

Selection of Members. — The members CE and HB are in 
tension and having about the same stress acting in them will be 

made of the same section. The net area required is **-*** — = 

12,000 

4.37 sq. ins. 

Trying two 4 in. X 3 in. X iV m - angles, and assuming J -in. 
diameter holes for f-in. rivets, the net area afforded by these 
angles is 5 sq. ins. Owing to their lengths their secondary stresses 
as given by the formula for combined stresses, Chapter VII, will 
be low and they will do. 

The member CF will also be made of the same section. 

Member GH. — The maximum stress for this member occurs 
when the load is at apex 2. The stress, given by Fig. 202, is 

27,000 lbs. The net area required then is — ■ = 2.2^ sq. ins. 

" H 12,000 ° H 

Trying two 3 in. X 3 in. X i-in. angles, which have an area 

of 2 X 1.44 = 2.88 sq. ins., and allowing one f-in. diameter hole 

for a f-in. diameter rivet to an angle the net section for two 

angles is 2.88 — (2 X 0.22) = 2.44 sq. ins., and these angles 

will do. 

Members AG and AH. — These members will be subjected to 

combined compression and flexure, and as it is difficult to brace 

these members laterally, due consideration must be given to the 

reduction of the fiber stress to provide for the column action of 

the compression flange. The axle load is — l = 5300 lbs. 

2 

The maximum bending on the 10-ft. span will occur when one of 



174 GRAPHICS AND STRUCTURAL DESIGN 

the wheels is one-fourth the distance between the wheels from 
the center of the span. 

Finding the reactions we have 

* 2 . (5300 X 54) + (5300 X 78) = 8 lbs _ 

I20 

Theni^i = 10,600 — 5830 = 4770 lbs. 

The maximum bending moment on the span then is 

M = i?i X 54 = 4770 X 54 = 257,580 in. lbs. 

The maximum stress will occur in these pieces in member HA 
when the load is located as shown in Fig. 204, and the direct 



'"-irl 



78^'- — 



Fig. 204. 



3 



stress is given by Fig. 203. To determine this stress the reaction 
at apex 1 must be found when the load is in the above position. 

Under this loading the apex load at 1 is R 2 = 5830 lbs. plus the 
equivalent frame load at 1, which is 750 lbs., these making a 
total of 6580 lbs. 

The direct stress is found to be 31,000 lbs. 

The jib will be assumed as braced in two places, so that it will 
be approximately braced laterally at intervals of about 8 ft. 

This bracing can be only a light frame, generally made of a 
bent angle that must clear the trolley. The total bending on 
the member AH is due to the vertical loading used in deter- 
mining the 257,580 in. lbs. just calculated, and in addition to 
this there is bending due to the hoisting-rope pull and the rack- 
ing-rope pull, both of which act to produce bending of the same 
character on the beam and must therefore be added to the bend- 
ing moment just found. 



CRANE FRAMES 175 

The bending moment due to hoisting-rope pull = 5650 X 19 = 
107,350 in. lbs. 

The bending moment due to the racking-rope pull = 
1000 X 9 = 9000 in. lbs. 

The total bending moment then is 257,580 + 107,350 + 
9000 = 373,930 in. lbs. Assuming that a section will be tried 
whose flange width is 3 ins., the ratio of unsupported part of 

span to flange width is ■ =32. 

o 
From the curve on page 102 the allowable fiber stress in com- 
pression for this ratio of length to flange width is about 85 per 
cent of the maximum desired. The allowable fiber stress is 
12,000 X 0.85 = 10,200 lbs. per sq. in. If we try 12-in. channels 
at 25 lbs. per ft., we find their area to be 2 X 7.35 = 14.7 sq. ins. 
and their section modulus 2 X 24.0 = 48.0. 

j OOO 

The direct stress per square inch is — = 2100 lbs. 

14.7 

The stress permitted for flexure becomes 10,200 — 2100 = 
8100 lbs. per sq. in. The required section modulus then is 

I = K = 373,930 = 

e f 8100 4 " ' 

It is evident, therefore, that the channels tried are satisfactory. 
Before being accepted finally, however, these channels should be 
examined as columns when the load is at its maximum radius. 
Under this condition the jib is subjected to a direct stress of 

s8,6<;o lbs. The unit stress due to this direct load is 5 ' 5 = 
* 14.7 

4000 lbs. 

The allowable unit stress according to Ritter's formula is 

- _ 12,000 _ 12,000 _ 12,000 

10,000 \rJ 10,000 

= 4900 lbs. per sq. in. 



176 GRAPHICS AND STRUCTURAL DESIGN 

The -value of 120 was found by dividing the unsupported 

length, 96 ins., by the radius of gyration of the channels about 

their minor axis, 0.78, or-^— = 120. 

0.78 

The 12-in. channels were found satisfactory in this respect. 

Mast. — The direct stress as given by the diagram, Fig. 201, 

is approximately 50,000 lbs. The reactions at the pins are 

^ 11 ,350 X 20 

R = — ^^ — - = 11,120 lbs. 

20.5 

The bending moment due to the above reaction is 

M = 11,120 X 21.5 = 239,080 in. lbs. 

Trying two 12-in. channels at 20.5 lbs. per ft., their area is 

Cd 000 
2 X 6.03 = 12.06 sq. ins., and the direct compression is — — — 

12.06 

= 4150 lbs. per sq. in. Assuming the allowable ultimate fiber 

stress in bending as 12,000 lbs. per sq. in. the balance for bending 

is 12,000 — 4150 = 7850 lbs. 

The required section modulus is 

i" M 2^9,080 

e f 7850 3 5 

An inspection of the tables of the properties of channels in a 
Manufacturer's handbook or on page 16 will show that either 
a 10-in. channel at 20 lbs. or 12-in. channels at 2o| lbs. will carry 
this load. However, since 12-in. channels are used for the 
member AG and as these channels give considerably greater 
stiffness they will be used rather than increase the number of 
sections demanded. As it is possible to brace the channels 
forming the mast at as frequent intervals as desired no account 
has been taken of the ratio of span to flange width. 



CHAPTER XIII 

GIRDERS FOR OVERHEAD ELECTRIC TRAVELING 

CRANES 

Specifications. — Design a bridge for a io-ton O.E.T. crane 
whose span is 45 ft. Assume the distance center to center of 
trolley wheels as 4 ft. o ins., that the bridge weighs 13,000 lbs. 
(two girders), and that the trolley weight is 6000 lbs. Design the 
girders to resist an additional loading laterally of one-tenth the 
live load. Cranes operating in the open may be subjected to 
lateral stress from wind pressure. Considering that during ex- 
treme wind storms the crane will probably not be kept in service, 
the bridge may be considered, firstly, as acted on by a dead load 
and the maximum wind load; secondly, as acted on by the dead 
load, the maximum live load with the attendant lateral stresses, 
and a minimum wind load. 

The maximum wind pressure may be taken at 30 lbs. per sq. 
ft. and is even taken as high as 40 lbs. per sq. ft. The assumed 
minimum wind pressure will vary with the conditions of the 
crane's location and with the judgment of the designer from 5 
to 15 lbs. per sq. ft. of exposed surface. 

Where a vertical bracing girder parallels the main girder the 
wind is assumed as acting simultaneously upon both girders. 
Allow a working fiber stress in tension of 10,000 lbs. per sq. in. 
and the same properly reduced by a column formula for com- 
pression pieces. Use f-in. rivets throughout. 

The loading will be reduced to that for a single girder or one- 
half the bridge. The weight of the bridge will be assumed as 
a uniform load and, as is usually done, the bending moment for 
it will be calculated for the middle of the bridge. 

Weight of 1 girder = — = 6500 lbs. 

2 

177 



i 7 8 



GRAPHICS AND STRUCTURAL DESIGN 



W -l 12 

Dead-load bending = — — = 6500 X45X -r* = 438,750 in. lbs. 

8 8 

Assuming the load as hung centrally on the trolley, the load on 

each trolley wheel will be 

J (trolley weight + live load) = J (6000 + 20,000) = 6500 lbs. 

The reaction 

Ri = [(6500 x 19.5) + (6500 x 23.5)] = 62IO lbs> 

45 
The maximum bending due to live load then is 

M = 6210 X 21.5 X 12 = 1,602,180 in. lbs. 
The total bending due to combined dead and live loads is 
M = 438,750 + 1,602,180 = 2,040,930 in. lbs. 

The material will be determined for the two following sec- 
tions, the depth being taken as 3 feet. 



. -21.5- - 



it- 



-^rrt 



-22to l 



H. 



fHl 



lL 



Fig. 205. 



Fig. 206. 



Fig. 207. 



The box section (Fig. 207) will be considered first. If the 
width is made one-thirtieth the span it will be, say, 17 ins. wide. 
The ends of the girders resting upon the bridge trucks will be 
assumed 18 ins. deep. The web plates will be made \ in. thick, 
making the gross web area at the ends 2 X j X 18 = 9 sq. ins. 

The maximum end shear is given by the diagram of maximum 
shears (Fig. 208), and is 15,800 lbs. The unit shearing fiber 



GIRDERS FOR TRAVELING CRANES 



179 



stress is 



15,800 



1760 lbs. This fiber stress is very low but, 



notwithstanding that some specifications permit yV m - material, 
we will use the J-in. plates. 




Fig. 208. 

Compression Flange. — The distance between the centers of 
gravity of the flanges will be assumed as approximately the 
distance back to back of the flange angles, or 36 ins. According 
to the curve, page 102, the allowable fiber stress in a compression 
flange when the ratio of the laterally unsupported length of 
flange to flange width is 30 to 1, is 80 per cent of the maximum 
desired. The allowable fiber stress becomes 10,000 X 0.80 = 
8000 lbs. per sq. in. 

The net flange area then becomes (see page 149), 

M 

7.1 sq. ms. 



2,040,930 
8000 X 36 



fXh 

If the web is considered as taking bending in addition to shear 
the web will furnish a section equal to one-eighth the web area, or 
I X 36 X i X 2 = 2.25 sq. ins. In this case the flange plates 
will work out so fight that the web plates will not be assumed as 
resisting bending. The flange angles will be tried at 3 ins. X 
3 ins. X i in. and they will be considered as having two holes, one 
in each leg at the same section; this makes it possible to locate 



180 GRAPHICS AND STRUCTURAL DESIGN 

the rivets at any points desired. The plate will be taken as 
17 ins. wide. Net area of angles, 3 ins. X 3 ins. X \ in., having 
two f-in. diameter holes, 

(2 X 1.44) — (4 X 0.22) = 2.88 — 0.88 = 2.00 sq. ins. 

The net area of the plate then is 7.1 — 2.00 = 5.10 sq. ins. 
The net width of the plate is 17 — (2 X J) = 15.25 ins. The 

thickness of the plate is — — = 0.334 in.; use y 5 g in. 

Tension Flange Area. — Here the allowable fiber stress is 
10,000 lbs. per sq. in. and the required net area of the flange is 

A M 2,040,930 , 

A = - = — ' ^ ' vo r = 5.67 sq. ins. 

fXh 10,000 X 36. ° ' H 

Using two, 3 in. X 3 in. X J-in. angles as before, the net area of 

the plate is 5.67 — 2.00 = 3.67 sq. ins. and the required thickness 

^ 67 
of the plate is = 0.241 in., say } in. • 

15-25 
The girders will be made approximately for uniform strength * 
(see Fig. 209) . The depths at points along the girder may be 




checked by drawing the diagram of maximum bending moments 
similar to that drawn for a railway girder, in Chapter XI, and 
combining the live-load bending moments thus found with the 
dead-load bending moments. 

Since M = flange area X mean fiber stress X girder depth, 
it follows that the flange area and the mean fiber stress being 
made constant the depths at any section can be made to vary 



GIRDERS FOR TRAVELING CRANES 



181 




182 



GRAPHICS AND STRUCTURAL DESIGN 



with the bending moments. This, of course, is approximately 
true only so long as the flange area times the square of its dis- 
tance from the neutral axis of the entire girder section at the 
point considered is large compared with the inertia of the flange 
about a parallel axis through its center of gravity. Where 
greater accuracy is desired the moments of inertia of the several 



sections may be computed and the formula M 



f - used. 
e 



The section chosen should now be tested when subjected to the 
lateral moment stated in the specification. This moment being 
produced by one-tenth of the live load it will equal one-tenth of 

the live-load moment, or — = 160,218 in. lbs. The web 

10 

plates for a distance of 1 2 ins. from the top will be considered as 



~J 



1 



-\ 



VaI Plate/ 



%s Plate 



IpYWk 



Fig. 211. 



resisting this moment. The following calculation gives the 
moment of inertia about the axis B-B, Fig. 211. 



bd* 



*r 



One plate, — ■ = 0.31 X — = 126.00 

12 12 

Two plates, Ah 2 = 2 X 0.25 X (12 - 0.88) X (5.12) 2 = 145-73 

Two angles (about their own axis) = 2.48 

Two angles, Ah 2 = 2 X 1.44 X 6.09 2 = 106.82 

Inertia 381.93 
Deducting for rivet holes (Ah 2 portion only), 

2 X 0.22 X S-38 2 = 12.73 

2 X (0.27 + 0.22) X 6.75 2 = 44-65 57-38 

Inertia 324.55 



GIRDERS FOR TRAVELING CRANES 183 

The extreme fiber stress is then / = — - — 

8 ? 
or / = 160,218 X — ^7 = 4200 lbs. per sq. in. 

324.6 

The mean fiber stress in the compression flange is 8000 lbs. per 
sq. in. 

The total fiber stress is 8000 + 4200 = 12,200 lbs. per sq. in. 

This fiber stress, considering that it is the maximum of both 
vertical and lateral bending, is satisfactory. 

Some designers prefer to estimate the forces acting laterally 
on the crane upon the assumption that the crane when carrying 
its maximum load is stopped from full speed in an assumed dis- 
tance, the retardation being considered uniform. In the present 
instance we will assume a velocity of bridge travel of 300 ft. per 
minute, and that the crane is stopped in 5 ft. 

The crane velocity being 3 ^- = 5 ft. per second, the retarda- 

v 2 
tion is/ = , where/ = acceleration or retardation in feet per 

2 • o 

second per second. 

v = velocity of bridge in feet per second. 

5 = distance traveled during acceleration or retardation in 
feet. 

In our case the retardation is f = — - — = 2. < ft. per sec. per sec. 

J 2 X 5 

As the force equals mass times retardation, the force required to 

stop the bridge is F = — X 2.5 = 1000 lbs. 

32.2 

This acts as a uniform load laterally along the bridge and will 

be resisted by both top and bottom flanges of each girder. The 

force to retard the trolley and live load will act upon the upper 

flanges of the two girders of the bridge and equals F = — l 

32.2 

X 2.5 = 2020 lbs. 

The central load on one girder the equivalent of the above 

forces is ±-%°-± + ^_o = II35 lbs . 



1 84 GRAPHICS AND STRUCTURAL DESIGN 

The bending moment due to this central load is 

M = ■ ■ = 1135 X 45 X — = 153^25 m. lbs. 

4 4 

This moment differs so slightly from the moment found by the 
previous method that the girder dimensions should prove satis- 
factory viewed from either estimate. 

Flange Rivets. — Considering that the unit fiber stress in 
tension is 10,000 lbs. per sq. in. the allowable unit shearing fiber 
stress on the rivets, assuming about the same material, would 
be f X 10,000 = 7500 lbs. per sq. in. The bearing value would 
be taken at twice the shearing value or 15,000 lbs. per sq. in. 
The value of f-in. rivets in single shear and bearing in a J-in. 
plate is 2813 lbs. 

Assuming the girder depth at the supports as 18 ins., the dis- 
tance center to center of rivets vertically is 18 — (2 X 1.75) = 
14.5 ins. 

The maximum end shear will be assumed as transferred to the 
girder in a distance equal to its depth, or 18 ins. The vertical 

shear per inch of span is S\ = — ^— — = 880 lbs. 

18 

The horizontal shear per inch at this point is 

Vertical shear 



$2 = 



Distance center to center of rivets vertically 



115,800 „ 

$2 = = ioqo lbs. 

14.5 

The resulting shear per inch of span is 

S3 = ^/si + s 2 2 = 1400 lbs. 

Considering that the web being double will necessitate a row of 

rivets in each web plate the horizontal distance center to center 

, . . 2813 X 2 

of rivets = = 4 ms. 

1400 

Owing to the concentration of loading at the ends the spacing 

will be made 3! ins. 



GIRDERS FOR TRAVELING CRANES 185 

The net area of the compression flange being 7.1 sq. ins. and 
the mean fiber stress 8000 lbs. per sq. in., the number of rivets to 

develop the full flange strength is — — , about 20. 

2813 

These rivets placed in two rows and spaced 3! ins. center to 
center would require a distance of 3.5 X 11 = 38.5 ins. This 
would indicate that for about 48 ins. from the ends of the plate 
the rivets should be spaced about 3! ins. center to center and at 
the other sections the spacing in the compression flange should 
not exceed from 16 to 20 times the thickness of the outside plate. 
The spacing in the two flanges is usually made the same. The 
upper plate being j 5 g in. thick the rivet spacing should not 
exceed 5 ins. 

Angle Stiffeners. — Allowing 10,000 lbs. per sq. in. the area of 

^ L .„ Maximum end shear 1^,800 

the stiffeners = = -»" — = 1.58 sq. ins. 

10,000 10,000 

Using two angles the smallest permitted would be 2\ ins. X 
2\ ins. X I in., and their section 2 X 1.19 = 2.38 sq. ins., which 
is more than ample. 

Stiffeners, however, serve a double purpose, as the opposite 
inside stiffeners are riveted to a plate forming a diaphragm at 
the section and greatly stiffening the girder laterally. They 
frequently also serve to secure short channel or angle sections 
which pass across the flange and assist the flange plate in carry- 
ing the rail. These latter are commonly placed at each pair of 
vertical stiffeners, while the diaphragms are generally placed at 
alternate panels. 

The web plate should be reinforced where brackets or motors 
are attached to it and at these points handholes should be cut 
in the web plate on the opposite side of the girder to facilitate 
bolting the brackets and motor to the girder. 

Girder Design for Fig. 206 
Using the same data the girder will now be designed for the 
section (Fig. 206). 
The maximum combined live- and dead-load bending was 



1 86 GRAPHICS AND STRUCTURAL DESIGN 

found to be 2,040,930 in. lbs. The maximum end reaction pre- 
viously determined was 15,800 lbs. The girder will be assumed 
36 ins. deep, back to back of angles, at the center of the span, and 
20 ins. deep at the bridge trucks. The web will be taken T 5 g in. 

thick making the unit shearing fiber stress — — = 2550 

0.31 X 20 

lbs. per sq. in. 

The effective depth will be assumed slightly less than 36 ins. 
to allow for the centers of gravity of the channels falling towards 
the center of the girder, the distance being 36 — 1.5 = 34.5 ins. 
The net flange area in tension then will be 

. M 2,040,0^0 

A = - — - = = 5.92 sq. ms. 

fXh 10,000 X 34.5 

Trying one 9-in. channel at 13^ lbs. per ft., its area is 3.89 sq. 
ins., and two 3 in. X 3 in. X yV-in. angles, we have 

Sq. ins. Sq. ins. 

One 9-in. channel at 13! lbs. gross area 3.89 

Less two rivet holes in f-in. material, 2 X 0.22 =. 0.44 3.45 

Two 3X 3X i 5 6- in - angles, 1.78 X 2 = 3.56 

Less four rivet holes in j^-in. plate, 4X0.27=... 1.08 2.48 

5-93 

This net area is satisfactory. 

In the compression flange the fiber stress must be reduced to 
provide for lateral strength. The ratio of span to flange width 
if a 15-in. channel is tried is 45 X yf = 36. Reducing the unit 
working fiber stress to 75 per cent of the maximum desired 
permits a fiber stress of 7500 lbs. per sq. in. Again substituting 
in the formula 

M = AX/Xh. A=-^-= 2 '° 4 °^° = 7.88 sq. ins. 

fXh 7500 X 34-5 

Trying two 3 in. X 3 in. X T6" m - angles and one 15-in. channel at 
33 lbs. per ft., 

Sq. ins. Sq. ins. 

One 15-in. channel at 33 lbs 9.90 

Less two rivet holes, 2 X 0.88 X 0.40 = 0.70 9.20 

Two 3X 3X xVin. angles, 1.78 X2= 3.56 

Less four rivet holes 1.08 2.48 

Total area 11.68 



GIRDERS FOR TRAVELING CRANES 1 87 

This seems to be too large but before discarding it the combined 
stresses due to the lateral and vertical bending should be deter- 
mined. The rlexural fiber stress with this flange is 

, M 2,040,9^0 ,, 

f = = — = ^100 lbs. per sq. in. 

J A Xh 11.68 X34-5 

In the preceding girder section the lateral bending due to stop- 
ping the crane in 5 ft. was estimated at 153,225 in. lbs. The 
fiber stress, assuming that the bending is mainly resisted by the 
15-in. channel whose section modulus about its axis 1-1 is 41.7, is 

/ = — — = -** 1 — - = 3680 lbs. per sq. m. 
I 4i.7 

The combined maximum stress then is 

(- ) + 3680 = 10,480 lbs. per sq. in. 

Vo-75 / 

This is low considering that everything of importance is taken 
into account, but will be used, as calculations made for 12-in. 
channels indicate a fiber stress undesirably high. This section 
of the girder may now be checked by calculating its moment 

of inertia and using the formula / = — • 

Flange Riveting. — These calculations are similar to those 
made for the box girder. The rivet value in a i 5 6~ m - we ^ plate 
for f-in. rivets is 3316 lbs. As in the other case 

15,800 „ 

Si = — = 790 lbs. 

20 

15,800 , ,, 
s 2 = , — = 960 lbs. 
16.5 

The resulting shear per inch of span is 



S3 = vsi 2 + s 2 2 = V790 2 + 960 2 = 1240 lbs. 

and the rivet spacing of f-in. rivets is ||Jf = 2.67 ins. 

The stiffeners can, as in the preceding problem, be made the 
smallest allowable, say, 2^ in. X 2^ in. X J-in. angles. 



1 88 



GRAPHICS AND STRUCTURAL DESIGN 



Bridge Girders with Horizontal Stiffening Girders 

The following example is intended to illustrate this type of 
bridge. Span 60 ft. Girder depth f f = 5 ft. back to back of 
angles. The crane capacity is to be 20 tons (40,000 lbs.). The 
trolley wheel base is 6 ft. o ins. The bridge weight, two girders, 
is 30,000 lbs. The load on each trolley wheel is 

40,000 -f- 12,000 

4 



13,000 lbs. 



The live-load bending will be found as in the preceding cases 
and as follows: 

Ri = (13,°°° X 3i-5) + (13,000 X 25.5) . lb 

OO 

M = Ri X 28.5 X 12 = 12,350 X 28.5 X 12, 
M — 4,223,700 in. lbs. 



28.£ 



-25.5^ 



-»J < *)' 

1.5' V 



Fig. 212. 



In estimating the dead-load bending the girder weight will be 
assumed as a uniform load. The bending moment is 

,, WL 15,000 X 60 X 12 . „ 

M = —— = -^ = 1,350,000 in. lbs. 



The total bending is 4,223,700 -f- 1,350,000 = 5,573,700 in. lbs. 

Allowing a mean unit fiber stress of 9000 lbs. per sq. in. the 
flange area will be 

. M 5,573,7oo , 

A = — - 7 = D,J ' 0, ' — - = 10.65 sq. ms. 
/ . h 9000 X 58 D H 

The distance between the centers of gravity of the flanges has 
been taken 58 ins. as probably no flange plates will be required. 



GIRDERS FOR TRAVELING CRANES 



189 



The web plate will be assumed as J in. thick and one-eighth the 
web area will be considered as contributing to the flange section. 

Web area = 60 X \ = 15 sq. ins. 

\ X web area = -* ¥ 5 - = 1.88 sq. ins. The balance of the flange 
area will be made up of the angles, or 10.65 — 1.88 = 8.77 sq. 
ins. 

The flange having two angles each angle must have a net area 

8 77 
of - lLL = 4.38 sq. ins. Allowing for two holes for f-in. diameter 

2 

rivets, the gross area of each angle must be 4.38 + 0.44 = 4.82 
sq. ins. 

Trying 6 in. X 4 in. X §-in. angles, their gross area is 4.75 
sq. ins., which is satisfactory. The flange will therefore be made 
of two 6 in. X 4 in. X f-in. angles. 

Horizontal Stiffening Girder (Fig. 213). — First assuming a 
horizontal loading of one-tenth the vertical loading gives two 



Fig. 213. 




© 



concentrated loads of 1300 lbs. each and a uniform load of 1500 
lbs. Before using these figures we will check them by estimating 
the forces acting on the girders when the crane carrying its full 
load at maximum velocity is stopped within 5 ft. 

The velocity of the crane being assumed at 300 ft. per minute, 
or 5 ft. per second, to stop the crane in 5 ft. the retardation must 



190 



GRAPHICS AND STRUCTURAL DESIGN 



be / = - — - = — - — = 2.5 ft. per sec. per sec. The force to 
2X5 



2 • s 



check the live load and trolley is Force = Mass X Acceleration. 

£, 52,000 X 2.5 „ 

F = s2 - J J = 4040 lbs. 

32.2 

The force to check the bridge is 

„ 30,000 X 2.5 „ 

F = — * = 2330 lbs. 

32.2 

The first ^ 4 -- = 1 010 lbs. compares with a concentrated load of 
1300 lbs., while the latter is a uniform load of 1165 lbs. compared 
with one of 1500 lbs. The first figures, being the greater, will be 
used. 

As this lateral bending is at best only a rough approxima- 
tion it will be sufficiently accurate to assume a central load of 
2 X 1300 = 2600 lbs., due to trolley weight and five load, and 
an equivalent central load of —4— = 375 lbs., due to bridge 

(2600 ~\~ "37 ^\ 
^-^ 1 = 1490 lbs. 

The stress AI, Fig. 214, is 8940 lbs. 

A section through the main and bracing girders is shown in 
Fig. 215. 

T 



i. 




Fig. 215. 



Fig. 216. 



As the wheels travel along the flange of the main girder the 
force of 1300 lbs. transferred to the girder by a wheel may pro- 
duce bending on this flange; for instance, between the two points 
of lateral support 1 and 2 this bending will be 

™ W-L 60 . „ 

M = = 1300 X — = 19,500 in. lbs. 

4 4 



GIRDERS FOR TRAVELING CRANES 191 

The moment of inertia of two 6 in.- X 4 in. X J-in. angles back 
to back but separated by a J-in. plate is 

Angles about their own axes, 2 X 1740 = . 34.80 

A - h 2 = 2X 4-75 X (1.99 + 0.12) 2 = 42-27 

/ = 77.07 

The fiber stress due to bending is 

- M -e 19,500 X 6.12 „ 

/1 = -— = vo = 1550 lbs. per sq. in. 

I 77.07 

Direct fiber stress due to horizontal bracing girder 
f 2 = -, — — r = 040 lbs. per sq. in. 

n (2 x 4.75) 

The total fiber stress then is 

9000 + 940 + 1550 = 11,490 lbs. per sq. in. 

As all the forces acting on the girder both vertically and hori- 
zontally have been considered this fiber stress is satisfactory. 
The unbraced length of AI being 60 ins. the radius of gyration 

if - = 120 then r = = - . The smallest angles fulfilling these 

r 120 2 

requirements are 3 ins. X 3 ins. Trying one 3.5 in. X 3.5 in. X 

A-in. angle, its least radius of gyration is 0.69. - = — — = 87. 

r 0.69 

According to Ritter's formula, 



1 1 ,0 00 j. _ 11,000 

8? 



~TJR' 

10,000 \r/ 



1+-^-- 1+ 



The total load on a 3.5 in. X 3.5 in. X iVin. angle is 6250 X 2.09 
= 13,000 lbs. This satisfies the requirements. 

The diagonals will carry their maximum stresses when the 
trolley fully loaded is at one side of the bridge. The maximum 
end shear on the horizontal girder is 

„ 2600 X 57 . O 1U 

Ri = ^-^ + 375 = 2845 lbs. 



192 GRAPHICS AND STRUCTURAL DESIGN 

The force in the diagonals then is 2845 X 1.41 = 4010 lbs. 

The length of the diagonal is 60 X 1.41 = 84.6 ins. If- = 120 

r 

then r = -^- ~ 0.70. This will require 3 J in. X 3^-in. angles. 
120 

The permissible fiber stress, according to Ritter's formula, is 

. 11,000 11,000 .. 

/ = 7772 = J" = 45°° lbs - P er sq. m. 

i+— — X 



© 



The load that can be carried by one 3I in. X 3J in. X T6- m - 
angle is 4500 X 2.09 = 9400 lbs. 

No calculation will be made for the vertical latticed girder as 
it merely holds the horizontal girder in place; the smallest per- 
missible angles, 2 J ins. X 2 J ins. X \ in., will be used. The lower 
horizontal girder will be made the same as the upper, except- 
ing that the angles at right angles to the main girder will be 
omitted. 

Riveting. — To determine the rivet spacing in the main 
girder it will be necessary to estimate the maximum shears along 
the girder. The maximum reactions on one girder are, live- 
load reaction, 26,000 X f -J = 24,700 lbs. The dead-load re- 
action is — — - = 7500 lbs. The combined live-load and dead- 
load reactions are 24,700 + 7500 = 32,200 lbs. See Fig. 217, 
which is the diagram of maximum shears for this girder. The 
maximum wheel load will be assumed as transferred to the web 
in 24 ins. of flange length, or the shear per inch of span due to 

11 , ,- 13,000 ,, 

load concentration is -*" = 540 lbs. = S\. 

24 

The horizontal shear at the ends per inch of span is the 

end shear divided by the vertical distance center to center of 

n • 32,200 Q2,200 _ „ 

flange rivets = =- Q -y rr = a - J = 585 lbs. = s 2 . 

* [60 - (2 X 2.5)] 55 



53 = vV+^T 2 = V540 2 + 5 8 5 2 = 796 lbs. 



GIRDERS FOR TRAVELING CRANES 



193 



The rivet value of a f-in. rivet in double shear, bearing in a 
f-in. plate, is 2813 lbs. for a unit shearing stress of 7500 lbs. per 
sq. in. and 15,000 lbs. per sq. in. in bearing. The rivet spacing 
therefore will be -■?■$■$- ~ 3I ins. 




Fig. 217. 

The rivet spacing will be determined for quarter points along 
the girder (see Fig. 218). At 15 ft. from the end the maximum 




shear is 22,000 lbs. and the horizontal shear per inch of span is 
22,000 



55 



= 400 lbs. 



s s = V400 2 + 540 2 = 672 lbs. 



194 GRAPHICS AND STRUCTURAL DESIGN 

The rivet spacing is -YtV" = 4.18 ins. 

At the middle the shear is 11,700 lbs. and the horizontal shear 

per inch of span is — — — = 213 lbs., hence 



s 3 = V213 2 + 540 2 = 580 lbs. 

The rivet spacing is -Wo 3 - = 4.85 ins. 

From the diagram, Fig. 218, the rivet, spacing for any point 
along the girder is readily scaled. 



CHAPTER XIV 
REINFORCED CONCRETE 

Reinforced concrete is designed by an extension of the 
principles applied to the discussion of structural steel. The 
modification is necessary since reinforced concrete is a compos- 
ite of several materials and therefore, unlike steel, is nonhomo- 
geneous. 

In ordinary practice the concrete varies from i, 2, 4 (1 cement, 
2 sand and 4 stone, slag or gravel) to 1, 3, 6. The best mixture 
depends upon the character of the materials, the purpose being 
to have the cement fill the voids in the sand, while the mortar 
thus produced shall fill the voids in the stone, slag or gravel. 

Mr. N. C. Johnson, in the Engineering Record of Jan. 23, 
Feb. 6, 13 and 27, March 6 and 13, 1915, discusses the nature 
of concrete as revealed by a microscopic study. He draws at- 
tention to the fact that all concretes are very much weaker than 
the weakest of their constituents and explains this by showing 
how far the commercial mixing of the cement, sand and stone 
varies from that theoretically desirable. 

The presence of air and water voids in the concrete prevents 
its attaining its maximum density and consequent strength. 
The concrete also contains unhydrated cement in considerable 
quantities which is revealed by the microscope; this also prevents 
its attaining its maximum strength. 

His researches showed that concrete exposed to water, either 
fresh or salt, deteriorated due to the capillary action of cracks 
in the concrete drawing water into it; this water produced in- 
ternal stresses by mixing with any unhydrated cement and pos- 
sibly also by depositing salt crystals within the concrete. 

In a concluding statement, Mr. Johnson estimates that less 
than 20 per cent of the cement added to concrete is used effec- 

195 



196 GRAPHICS AND STRUCTURAL DESIGN 

tively. It is to be hoped that the final results of such investi- 
gations may lead to a commercial method whereby a concrete 
of maximum density may be readily obtained. 

Cement. — Portland cement only should be used. This is 
practically the only cement available, as very little natural 
cement is now manufactured. 

The cement should meet the usual standard specifications. 

Sand. — The sand should be a clean, coarse, sharp sand free 
of clay and dirt. Fine, rounded or dirty sand greatly reduces 
the strength of the concrete. 

Stone and Gravel. — Stone is usually broken so that the 
maximum size does not exceed from f in. to if ins. Stone 
and gravel should be freed from dust, sand and dirt, but uni- 
formity of size is not desirable as it increases the percentage 
of voids. 

Mixing. — Practice formerly demanded a fairly dry mixture, 
tamped until the water came to the top, but present practice 
uses a very wet mixture, which greatly reduces the tamping 
required and makes a more solid concrete. The wet concrete 
is weaker than the other when new, but its strength rapidly 
approaches that of the other with age. 

Physical Properties. — When one month old and made from 
ordinary materials the compressive strength of 1, 2, 4 stone 
concrete can be assumed at from 1800 lbs. to 2200 lbs. per sq. in., 
while a 1, 3, 6 mixture will have a strength of about 20 per cent 
under these figures. 

The modulus of elasticity increases with age and decreases 
with the unit stress, so that only approximate values can be 
assumed. A commonly accepted modulus of elasticity for mak- 
ing reinforced-concrete calculations is 2,000,000 lbs. per sq. in., 
being one-fifteenth that of steel. 

Elastic Limit. — There is no true elastic limit for concrete 
as it shows permanent deformation under the lightest loads. 
According to Bach and others it seems fair to assume that what 
is generally accepted as the elastic limit in concrete occurs at a 



REINFORCED CONCRETE 



197 



unit stress of from one-half to two-thirds of the ultimate com- 
pressive strength. 

PHYSICAL PROPERTIES OF STONE OR GRAVEL CONCRETE 





Age, days. 


1, 2, 4 Mixture. 


1, 3, 6 Mixture. 


Compressive strength .... 

Tensile strength 

Shearing strength 

Modulus of elasticity 

Coefficient of expansion, 
i° F 


30 
30 
30 
90 


1,800-2,200 

150-200 

1,000-1,400 

2,500,000-3,500,000 

O . OOOO060 
I50 


1,450-1,750 
IOO-125 
900-1,100 
2,500,000-3,500,000 

O . OOOO060 


Weight per cubic foot in- 
cluding steel 




i$o 







Note. — All stresses in pounds per square inch. Slag or cinder concrete will 
only develop about one-third the strength of a good stone concrete. Its modulus 
of elasticity will range from 1,000,000 for lean mixtures up to 2,500,000 for 1, 2, 4 
and richer mixtures. It weighs about 120 lbs. per cu. ft. 



Usual Unit Working Stresses. Flexure 



Extreme fiber stress on concrete, compression . . 

Extreme fiber stress on concrete, tension 

Steel, soft, tension 

Steel, mild and hard, tension 

Bonding stresses, straight bars 

Bonding stresses, straight bars, over supports . . 

Bonding stresses, bent or twisted bars 

Shearing, plain web, no steel stirrups 

Shearing, when reinforced with stirrups, or bars . 

Tension, diagonal, on concrete . , 

Modulus of elasticity of concrete 



Lbs. per sq. in. 

500-650 

o 

12,000 

16,000 

60-80 

90-1 20 

150 

30 

100 

30 

2,000,000 



Modulus of elasticity of steel ■ 30. 



000,000 



Usual Unit Working Stresses. Columns 

Lbs. per sq. in. 

Columns not reinforced . 450 

Columns, reinforced longitudinally only 450 

Columns, reinforced with hoops or bands 550 

Columns, reinforced with hoops or bands, and with from 1 

to 4 per cent of longitudinal steel 650 

Columns, structural steel thoroughly inclosing concrete 650 

In general the following nomenclature and formulae conform 
to the report of the Joint Committee on Reinforced Concrete of 
the American Society of Civil Engineers, American Society of 



198 GRAPHICS AND STRUCTURAL DESIGN 

Testing Materials, American Railway Engineering and Main- 
tenance of Way Association and the American Portland Cement 
Manufacturers. 

/. = tensile unit stress in steel, pounds per square inch. 

f e = compressive unit stress in concrete, pounds per square 

inch. 
E 8 = modulus of elasticity of steel, pounds per square inch. 
E c = modulus of elasticity of concrete, pounds per square inch. 

E c = n ' 

M = moment of resistance or a bending moment, inch pounds. 
A = area of steel, square inches. 

b = width of beam, inches. 

d = depth of beam to center of reinforcement, inches. 

k = ratio of depth of neutral axis to effective depth, d. 

x — depth of resultant compression from top. 

j = ratio of lever arm of resisting couple to depth, d. 
j • d = d — x = arm of resisting couple. 

p = ratio of steel to concrete. For rectangular beams, 

B = width of flange of T beams, inches. Here p = — — -• 

3 • a 

b 1 = width of stem of T beams, inches. 
t = thickness of flange of T beams, inches. 

z = U X E c 
fcXE 8 

Shear and Bond Stresses 

V = total shear, pounds. 

v = shearing unit stress, pounds per square inch. 

u = bond stress, pounds per square inch of bond area. 

= circumference or perimeter of bar, inches. 
So = sum of the perimeters of all bars, inches. 
T = allowable tensile load in bar or stirrup, pounds. 



REINFORCED CONCRETE 1 99 

Beams with Double Reinforcement 

A 1 = area of compression steel. 
p 1 = steel ratio of compression steel, not percentage. 

A 1 

. p 1 = — 

F BXd 

fs 1 = unit compression in steel. 

C = total compression in concrete. 

C 1 = total compression in steel. 

d 1 = depth to center of compressive steel. 

e = depth to resultant of C and C 1 . 

-s- 

Columns 

A = total net area. 
A 8 = area of longitudinal steel. 
Ac = area of concrete. 

P = total safe load. 

Theoretical Discussion of Reinforced-concrete Beams 

Rectangular Beams. — The usual formulae for the design of 
reinforced-concrete beams are developed upon the assumptions, 
(a) that a cross-section plane before bending continues a plane 
during bending, and (b) that stress and strain are proportional. 

From these assumptions the strain in compression a unit dis- 
tance above the neutral axis must equal that in tension a unit 
distance below that axis, or, from Fig. 219, 

kd XE C = (d - kd) E 8 = (jd - f kd) E s = d{j-%k)Ej 
f 8 X E c jd -jkd d - kd _ 
f c XE s ~ kd ~ kd " *' 

from which k = > (1) 

s + 1 

and 1 = -^ — ! — r • (2) 

J 3 (» + 1) 



200 



GRAPHICS AND STRUCTURAL DESIGN 



According to the principles of mechanics the sum of the hori- 
zontal forces acting on a yertical section must be zero, hence the 
resultant force in the compression flange must equal the result- 
ant force in the tension flange, or 

A Xf s =t c Xk.dXb, 

2 

and the moments are also equal, M 3 = M c . 
M 8 = A Xf s Xj-d. 

M c = ^Xk.dXbXj>d=^Xk.jXb.d 2 . 



(3) 
(4) 




I / Jed 

f-r 




i- 



These equations apply when the maximum fiber stresses f c 
and f s are known; the reinforcement then is the critical ratio 
for these fiber stresses. 

It is sometimes desirable to discuss the beam when a ratio of 
reinforcement is assumed which is not necessarily the critical 
ratio. 

From the fundamental equations, 

E s = n(i - k) 
k -E c k 



if = (1 - 



From equations (3) and (4), 

n(i — k) k , k 2 

— ^- = — and p = 7 rr 

k 2 p 2 »n (1 — k) 



hence 



(s) 



REINFORCED CONCRETE 201 

Solving for k we find 

k = V2 'p -n + (p -n) 2 — p '7i. (6) 

k 

The fiber stresses may be found by substituting these values 
of k and j in equations (3) and (4). 

Approximate Formula 

The curves on page 202 give the calculated relations between 
z, k,j, and n • p. It will be noted that for the usual values of p, 
j approximates f and k approximates f . The following approxi- 
mate formulae, using these values of k and j, are sometimes used. 

M s = lxAxf s Xd. (7) 

M c = iXfcXb-d 2 . (8) 

The curves, Fig. 220, give the relations between the several 

quantities. 

These curves are plotted upon a base line giving the values of z; 

f X E c 
therefore when z = ^f zr is known, the other values k, /, / • k, 

p • n and p are read on the perpendicular to z. Thus for z = 2.2 
the values are read on the dotted line a^a through 2.2. The 
values are j = 0.895; ^ = °-3°75 ^ *J = °- 2 75; p -n = 0.072 
and p for n = 15 is p = 0.0048. 

Had the value of p f or n ■— 15 been given as 0.0048 instead of 
the value of z = 2.2 the horizontal line b-b would have been 
drawn through p = 0.0048 until it cut the curve of p f or n = 15, 
and through this point of intersection the line a-a would be 
drawn perpendicular to the z axis and upon this line a-a the 
other values would be read as before. 

The following example will illustrate the use of the formulae 
and curves. 

Example. — A rectangular beam in which d = 22 ins. is to 
resist a bending moment of 384,000 in. lbs. Assume w = 15, 



202 



GRAPHICS AND STRUCTURAL DESIGN 




REINFORCED CONCRETE 203 

f 8 = 16,000 and f c = 650; find the width of the beam and the 
area of the steel. 

z _ /« x Ec = l6 >°°° = x 64 
foXE s (650X15) 

Interpolating from the curves, k = 0.379, j = 0.874, k *j = 

0.331 and p = 0.0077. Substituting these values in equation (4) 

gives 

, 2 M c 2 X 384,000 

= = — = 7 37 ins 

fcXk-jXd 2 650 X 0.331 X22 2 '* 0/ 

^4 = b X d X p = 7.37 X 22 X 0.0077 = I - 2 5 S <1- ms - 

The solution of this problem by the approximate method gives 
almost identical results. 

M = lxA.f a .d, 

,, ' A 8 -M 8 X 384,000 

therefore A = : — : = / = 1.25 sq. ms. 

7 -f.-d 7 X 16,000 X 22 ° H 

lf-§tf..».«> or >-£* 

, 6 X 384,000 

= — — = 7.32 ms. 

650 X 22 2 ' ° 

Another example will illustrate the procedure when it is re- 
quired to check an existing beam. A reinforced-concrete beam 
has the following dimensions: b = 9 ins., d = 20 ins., A = 1.50 
sq. ins., n = 15, and it is desired that f s should not exceed 
16,000 lbs. per sq. in. nor/ c exceed 650 lbs. per sq. in. 

A 1.50 

P = bTd = fo^) = °-°° 8 33- 

From equation (5), 

i s l a ^^ = 2, 4 

/c 2 £ (2 X O.O0833) 

and f 8 = 23.4 X 650 = 15,200 lbs. per sq. in. 

This fiber stress being under the 16,000 lbs. per sq. in. is satis- 



204 



GRAPHICS AND STRUCTURAL DESIGN 



factory. The bending moment corresponding to these fiber 
stresses can be found from either equation (3) or (4). 
M S = A XfsXj -d = 1.5 X 15,200 X 0.870 X 20 = 397,000 in. lbs. 
Checking this by the other formula gives 

M c = J (fc X k -j X b • d 2 ) = 396,000 in. lbs. 

This is about as close agreement as can be expected considering 
the curve readings, the difference being under 1 per cent. 

T Beams 

In reinforced-concrete design the section most frequently met 
is the T beam, shown in Fig. 221. The floor is usually a slab of 



:::n::::: 7^ fen 

■--1-1 r\ fc , U 




Fig. 221. 



reinforced concrete laid on floor beams, corresponding in the 
ordinary timber construction to joists. In the case of concrete 
construction the floor beam and the slab over it become one 
piece, the web stresses at the junction of the two being re- 
sisted by the concrete and usually by reinforcing stirrups or rods. 



REINFORCED CONCRETE 



205 



When the neutral axis 1-1, in Fig. 222, falls in the flange or 
slab the formulas and table previously derived for rectangular 




Fig. 222. 



beams are sufficient. When, however, the neutral axis falls in 
the stem, as is usually the case, another set of formulae is required. 



General Formula for T Beams Reinforced for Tension 

Only 

Neglecting the compression in the stem, and on the same 

conditions a and b that applied to rectangular beams we have as 

before : 

AXf ... f f = n(L^l). and , = _^_. 
fcXE a fc k z + 1 



Also from Fig. 223 



I* 

h 



k-d 



1 


1 


B 


J 

1 

1 


I 


' 1— 


A 
id 


1 1 
1 











:x 



=t= 



Fig. 223. 

Assuming that the fiber stress in the concrete varies as its dis- 
tance from the neutral axis, the distance x from the top of the 
beam to the resultant compression R is 



fc+fi 3 



3 • k • d - 2 -t x t 
2 • k • d — t 3 



2o6 GRAPHICS AND STRUCTURAL DESIGN 

this gives j-d = d-x = d- ( 3 ' k ' d ~ 2 f X -V ( 9 ) 

\ 2 • k • d — t 3/ ^ J 

The average fiber stress in the concrete is 

It follows that 

M c = BxtXf c (i -^jj'd 9 (10) 

and M a = A Xf 8 Xj -d. (11) 

Equation (11) can be approximated as 

M 8 = A *Js( d --\ 

When the ratio of the reinforcement is known, not the fiber 
stresses, equation (1) cannot be used to find k as z is not known. 
Taking equations (10) and (11), we find 

fc _ A 1 pd k 

2 kd \ 2 kd) 

npd + — 

It follows that k = — • 

t + np d 

This may also be written 

2 n dA + Bt 2 



k-d 



2 nA + 2 Bt 



To facilitate the solution of problems the following curves, 
Fig. 224, give the values of j and k for varying values of p and 

-and also the ratios of y for various values of k. These values 
d fc 

have been calculated with n = 15, this being a very commonly 

assumed value for rock concrete. For other values of n similar 

curves may be drawn or the formulae used. 



0.90- 



0.80 



0.70- 



0.60- 



0.50- 



0.40- 



0.30- 




0.20 



t01-2 



T— Beams 

n=l5 

Single Reinforcement 

M S =A x f s x j.d 

M^Bxtxfcd-^j.d. 



.014 
,016 




Scale of 



Fig. 224. 

Note. — The values 0.002 to 0.016 placed at the curves are the values 

A 
of p = — — -• Moving the decimal point two places to the right expresses 
B • a 

the reinforcement as a percentage. p. 207 



208 GRAPHICS AND STRUCTURAL DESIGN 

To find j if- = 0.25; from 0.25, at a on the upper scale, drop 

a perpendicular until it cuts the curve corresponding to the 
steel ratio of, say, 0.003 in b; project the point b horizontally 
until it cuts the vertical scale in c. Here the value of j is given 
as 0.914. To find k continue the vertical a-b until it cuts the 
curve of k corresponding to a steel ratio of 0.003 in d. This point 
d projected horizontally cuts the vertical scale in e or k = 0.262. 

To find the ratio of y produce the value of k horizontally until 

it cuts the curve of j in /. The vertical projection of / cuts the 

h 

f . f 

scale of ratios of y in g or y = 42.3. 

An example will further illustrate the uses of the formulae and 
curves. . 



H— — -f« 

H-^— 1 ST" 



1 tv 

1. 



if 

4-^'Square.Bais 



Fig. 225. 

Given n = 15,/c^ 600, f 3 = 15,000, and the following section, 
Fig. 225. Area of bars = 4 X 0.766 = 3.06 sq. ins. 

V06 t 4.5 

p = — = 0.0027. -; = = 0.237. 

r 60 X 19 d 19 

Consulting the curves, 

j = 0.918, k = 0.252, y =44-5- 

The resisting moment then is 

M 8 = A Xf 8 Xj'd 
or 

M 8 = 3.06 X 15,000 X 0.918 X 19 = 801,000 in. lbs. 



REINFORCED CONCRETE 2O0 

The fiber stress in the concrete is 

y = 44-5 = -^7 — or fc = 33° lbs. per sq. in. 

Jc jc 

Slabs 
In using the formulae and table for slabs it is customary to 
assume a width of slab of 12 ins.; thus assuming a slab whose 
span is 6 ft. and whose total dead and live load is 200 lbs. per 
sq. ft., taking the bending moment at 

Wl 

M = — , n = 15, f c = 600 lbs. per sq. in., f 8 = 1 2,000 lbs. per sq. 
10 

in.; the thickness of the slab and area of reinforcing steel can 
be found by the following calculations: 

... Wl (200 X 6) X 6 X 12 Q , . „ 

M = — = = 8640 in. lbs. 

10 10 

z = EcXf 8 = 12,000 = 

E s Xfc 15X600 vW " 

From the curves, page 202, the values corresponding to z = 
1.33 are k = 0.433,7 = 0.855, k 'J = °-37° and P = °- 011 ; sub- 
stituting these values in equation (4) gives 



A 4 / 2Mc . / 2 X 864O 

d = \f 7- : — = 1/ = 2.55 ms. 

V bXfcXkj V 12 X 600 X 0.37 

The area of the reinforcement then is given by equation (3) 
or by taking b • d • p = 12 X 2.55 X 0.011 = 0.337 sq. in. 

If f in. is allowed below the center of the reinforcing steel the 
slab thickness will be 2.55 + 0.75 ~ 3.25 ins. 

Bending Moments 
Owing to the monolithic construction of reinforced concrete 
the bending moments on beams are frequently estimated as 
lower than those occurring on similarly loaded but merely sup- 
ported beams. Thus for continuous slabs the moments are 

commonly taken as M = — , at the middle of the span and at 

10 

supports. 



2IO GRAPHICS AND STRUCTURAL DESIGN 

The rods inserted for slab reinforcement should be bent up 
near the \ points of the span and carried over the supports in 
the upper portion of the slab. Although some designers bend all 
the rods, it would seem sufficient and better practice to run from 
one-quarter to one-half of the rods through straight, only bend- 
ing the remaining bars. Although the angle of inclination of the 
bars should not be too slight it should preferably not exceed 30 
degrees with the horizontal. See Specifications, § 212. 

Square slabs reinforced in both directions and supported on 

four sides have the moments at the center of the span in one 

Wl 
direction taken as M = — . Beams and girders although some- 

20 

Wl 

times figured as fixed or continuous beams with M = — are 

10 

more commonly taken as supported beams with M — — . In 

8 

the above, W = the total load on the span in pounds and I = the 

span in inches. 

Beams with Double Reinforcement 

In the preceding discussions reinforcing steel has been assumed 
in the tension side of the beam. The case will now be considered 



Fig. 226. 

where the reinforcement is in both flanges. This condition 
occurs regularly in ordinary construction where some of the re- 
inforcing bars in the tension flange are bent up and carried over 
the support in the upper flange. In this case the beam resists a 
negative moment over the supports which creates tension in the 
upper flange and compression in the lower flange. In the case 



REINFORCED CONCRETE 



211 



of T beams the lower flange width being b the width of the stem 
is much less than the width B of the upper flange, and unless 
this narrow flange is reinforced for compression the fiber stress 
f c might easily be excessive. 

To meet the usual specifications (see Specifications, paragraph 
212), which demand that continuous beams be able to develop 
the same resisting moment at the supports and at the middle of 
the span, requires a tensile reinforcement at the supports equal 
to that at the center of the span if the beam is of uniform depth. 




Fig. 226a. 



As the bending moments reduce very rapidly near the sup- 
ports towards the center of the span, the forces acting in both 
tension and compression flanges are readily reduced by haunch- 
ing the column where the beam or girder runs into it. This 
amounts to increasing the depth of the beam adjacent to the 
column, as shown in Fig. 226a. 

The following theory applies to beams with double reinforce- 
ment. 

The nomenclature is that previously given, page 199. 

Assuming the compression in the concrete as varying propor- 



212 GRAPHICS AND STRUCTURAL DESIGN 

tionally to the distance from the neutral axis and that the con- 
crete carries no tension, we have, in Fig. 226, 

/. _ d - kd and // _ kd - e 
nf c kd njc kd 

Also A Xfs = C + C = A'j: + if c >kd-b, 

from which it follows that 



kd 
and 



= J in (A + <$>A')d f n(A f + A) i 2 _ n (A f + A) 



k = V2 n (p + <f>p') + n 2 (p + p'f - n(p + p'). 
To find the lever arm of the couple acting in the beam 

A Xf s Xjd = ^f c .kd.b(d--) + A'j: {d - e), 

J a Af. A-f. 

Expressing / c , /„ /,' and d in terms of k, <j>, p and p' gives 

3 np (1 - k) 

The relations between the fiber stresses are 

f _ M . , k , 

h AxjXd' Jc n (1 - £) 7 " 

and //= -(^) x/c = |^| /j . 

In the Engineering News of August 22, 191 2, Mr. Arthur G. 
Hayden gives the following approximate formulae ior j of rein- 
forced-concrete beams with double reinforcements. 

j = 0.93 + 2 p' + 100 ^' — 4p — 0.1 <f> — 20 //<£. 

Mr. Hayden states that the maximum error does not exceed 
1 \ per cent for all proportions of reinforcement between \ per 



REINFORCED CONCRETE 213 

cent and 2 per cent, and that for ordinary cases the error is in- 
appreciable. This formula assumes that n = 15. 

The following example will illustrate the use of the formulae. 

A T beam in which d = 21 ins. and b = 12 ins. is reinforced 
over the supports by two J-in. round bars in each flange. If f s = 
16,000 lbs. per sq. in., <f> = -j 1 -^ and n = 15, what moment can 
the beam resist at the supports and what are the fiber stresses 
/ c and//? 



= J 2n(A +<j>A')d + U 



kd ^ A+A ' y 



n{A + A r ) 
b 



/^o 1.2+— X21 
W = V /LA 1°1 JiiXMT iiXM, 

V 12 L12J 12 

kd = 5.85 ins. 

#g -!) + **' (!-*)(*-*) 

^ ~ np(i - k) 

o.28 2 (o.5o '—-\ + 15 X 0.0048(1.0 — 0.1) (0.28 — 0.1) 



J = 



15 X 0.0048 (1 — 0.28) 
j = 0.91 and jd = 21 X 0.91 = 19. n ins. 



By the approximate method^ = 18.97 ms - 

The bending moment, allowing /, = 16,000 lbs., is 

M = A Xf a Xjd = 1.2 X 16,000 X 19. n = 366,912 in. lbs. 

U = "73 ZjtU = — / ,oa X 16,000 = 410 lbs. per sq. in. 

n(d - kd) 15 (21 - 5.85) 

,, k — <f> , 0.28 — 0.10 w , „ 

/, = 7 /, = —- X 16,000 = 4000 lbs. per sq. in. 

1 — k 1 — 0.28 

The following problems are intended to be solved by means 
of the theory just given and it is recommended that they be first 



214 



GRAPHICS AND STRUCTURAL DESIGN 



done by using the formulae and then checked by the assistance 
of the curves. 

Problem. — The combined live and dead loads upon a floor 
slab are 140 lbs. per sq. ft.; assuming f c = 650 lbs., f 3 = 16,000 



E 

lbs. per sq. in., — s = 15, and that M 



Wl 



, determine the thick- 



ness of the slab from the top to the center of the reinforcement 
and the spacing of f-in. round bars if the span of the slab is 10 ft. 
8 ins. 

Problem. — Assume a continuous girder whose span is 25 ft. 
o ins., carries 50,000 lbs., d = 23 ins., B ~ 100 ins., slab thickness 

4 J ins., — - = 15, and M = — . Find the fiber stress in the 
E c 10 

concrete and steel if the steel reinforcement consists of two 

ij-in. and two if-in. round bars. 

Parabolic Variation of Stress in Concrete 

In the preceding discussion of reinforced-concrete beams it 
has been assumed that the fiber stress in the concrete above the 




d 1 — 



L 




Fig. 227. 



Fig. 228. 



neutral axis varies as the intercepts in the triangle; this is repre- 
sented by Fig. 227. 

Some designers prefer to assume that this stress varies as the 
intercepts in the parabola as shown in Fig. 228. The resultant 
flange force acts at a distance f k • d from the top of the beam 
and the average fiber stress is |/ c . This treatment is commonly 
limited to rectangular beams as it becomes too complicated 
when extended to T beams. The following formulae are derived 



REINFORCED CONCRETE 215 

similarly to those previously determined and apply to rectan- 
gular beams only. The symbols are those previously used. 

8* + 5 A u 8 / ^ 

1 = —, +■ and k = - (1 — ;). 

J 8(2 + 1) 3 

M 8 = A Xf s Xjd, 
Mc = ifcXb X kj X d\ 
Where the ratio of reinforcement is known the formulae are: 



k = V%p -n + (\p -n) 2 — \p -n and 7 = 1 - | k, 

JC 2b-kjd 2 

, _ M s 

Js 



p.b.jd 2 

The difference between these formulae and those previously 
given will be evident upon working several of the problems by 
both methods. Considering the character of the materials used 
in reinforced concrete it is a question whether methods intro- 
ducing any greater refinements than those used in the derivation 
of the first formulae are warranted. 

Most elaborate tests of reinforced-concrete beams have been 
made at the Experiment Station of the University of Illinois 
under the direction of Prof. Arthur N. Talbot, beginning in 1904 
and reported in the Bulletins of that station since then. Pro- 
fessor Talbot has deduced formulae and offers suggestions for 
reinforced-concrete design as the results of the tests; this data 
can be readily obtained by reference to the above bulletins. 

Web Stresses 

It is shown in the section of problems (see problems 65 to 68) 
that ordinarily in steel beams of usual sections the web stresses 
are of minor importance, while more careful consideration must 
be taken of them in timber beams, particularly of horizontal 
shear in timber beams of short spans. In reinforced concrete 
beams the consideration of the web stresses is of still greater 



2l6 



GRAPHICS AND STRUCTURAL DESIGN 



importance, and generally the webs of beams and girders must 
be carefully reinforced with steel to resist the web stresses. 

Taking section 2-2 in Fig. 229 the horizontal shearing force 
will increase from zero at the upper fibers to a maximum at the 
neutral axis 1-1; this force F 2 is then transferred to the hori- 
zontal reinforcing steel. Assuming that the bending at 3-3 ex- 
ceeds that at 2-2 the horizontal shearing force F 3 at 3-3 will 




Fig. 229. 

exceed F 2 , and if b is the width of the beam at the neutral axis 

the unit shearing force on the section b X 1 is v = -f -. 

X 1 

Equating the moments V X 1 = (F 3 — F 2 ) j • d or F z — F 2 = 



and v 



V 



j • d b »j *d 

It has been previously shown that j " • d ~ | d, hence 



v = 



0.875 & •& 

It should be noted that the reasoning is identical with that 
used in determining the flange riveting in plate girders. 

Tests of reinforced - concrete 
beams exhibit failures along lines 
a-a as shown in Fig. 230. This 
illustrates failure by diagonal 
tension. According to Prof. 
Arthur N. Talbot web failures 
of reinforced concrete beams 
occur at calculated shearing values much below the true 
shearing strength of concrete. Tests indicate these shearing 
strengths as ranging from 50 to 75 per cent of the compressive 




Fig. 230. 



REINFORCED CONCRETE 217 

strength. The usual method of web failure is by the tension 
resulting from the combination of the shearing and tensile web 
stresses. See Specifications, paragraph 2i8d. 

The beam is strengthened to resist this tension, first, by 
bending the reinforcing bars when no longer required in the 
flanges (see Fig. 239) ; secondly, by vertical stirrups (Figs. 239 
and 240) which carry the vertical component of the diagonal 
stresses, or thirdly, by a combination of vertical stirrups and 
bent rods, as shown in Fig. 239. 

If we consider any point in the body of the beam, it will be 
acted on by a unit horizontal tension or compression, a unit 
horizontal shear and a unit vertical shear. The resulting maxi- 
mum tensile or compressive stress is 



where f tm = unit maximum diagonal tension, 

ft = unit horizontal tension, 
v = unit shear, vertical or horizontal. 

The direction of the maximum tension is given by 

2 v 
tang 2 = — — > 

/* 
where f t = o, 6 = 45 degrees. It therefore follows that along 
the neutral axis and on the tension side of this axis 6 = 45 
degrees. 

Where 6 = 45 degrees, f t =o and f tm = v. 

It is seen then that the maximum diagonal tension acts at an 
angle of 45 degrees with the neutral axis and that the maximum 
unit diagonal tension equals the unit shear at that point. 

, V 

v = unit vertical shear. 

b = width of beam if rectangular. Use b\ the width of the 

stem if a T beam. 
V = total vertical shear on section. 



218 GRAPHICS AND STRUCTURAL DESIGN 

This formula applies for the worst condition, namely, that no 
tension is carried by the concrete. When it is desired to allow 
for the concrete the numerator of the fraction becomes V — 
v c • b *jd for rectangular beams and V — v c • b' *jd for T beams. 
Here v c = the allowable unit shear on the concrete. 

Let T = the total allowable tension in a diagonal bar, in 
pounds. If the bar makes an angle of 45 degrees with the hori- 
zontal, the spacing s along the horizontal in the first case is s = 

— — , while, when the allowance is made for the portion 

of the tension carried by the concrete, 

= 1.41 X T Xjd 

V — v c *b *.jd 

The factor 1.41 is introduced since the diagonal tension is 
measured along a 45-degree line, while the spacing is laid off on 
a horizontal line. Where the angle of inclination of the bent 
rod with the horizontal is a degrees instead of 45 degrees the 

factor 1. 41 should be replaced by — 

sin a 

To be effective for web reinforcement the spacing of diagonal 
bars should not exceed s = d. 

Vertical Stirrups 

When vertical stirrups are used they carry the vertical com- 
ponent of the diagonal tension. This is assumed as measured 
by the vertical shear. The spacing for vertical stirrups when no 
allowance is made for the shearing strength of the concrete is 

T ' j d 
s = - . Where a unit shearing stress v c is allowed in the 

concrete the spacing is 

T-jd 
s = • 

V — v c • b *jd 

In these formulae j may be ordinarily assumed at f without 
serious error. In the case of T beams b is the width of the stem. 
T is the total allowable tension in the stirrups, in pounds, s is 



REINFORCED CONCRETE 



219 



the stirrup spacing in inches, measured horizontally along the 
beam. This spacing should not exceed the effective depth of 
the beam, jd, and is commonly limited to from J to f the depth 
of the beam. 

In important beams and girders where the web tension is 
carried by vertical stirrups the necessary stirrup spacing may 



S) 



s\ 



/lf\ 



u 

Fig. 231. 



/I 



% 



Fig. 232. 



be determined similarly to the method employed for flange 
rivets in girders, on page 193, and as given by Taylor and Thomp- 
son in "Concrete Plain and Reinforced." Having determined 
by the formula the stirrup spacing at the end, the middle and 
several intermediate points on the span, lay off these spacings 




vertically upon a horizontal base line representing one-half the 
span, and at their proper positions, then draw a curve through 
these points. The required spacing can be scaled off for any 
point. 

If the same scale is used for laying off the half span and the 
vertical ordinates representing the spacing, then the succeeding 



220 GRAPHICS AND STRUCTURAL DESIGN 

positions for stirrups may be found by laying off ac = ab, 
ce = cd, eg = ef, etc. This is quite readily done by drawing 
be, de,fg, hi, etc., making angles of 45 degrees with the horizontal 
and alternately drawing the vertical and inclined lines. 

Bond Stress 

The diagram, Fig. 234, shows how the flange force varies from 
the supports to the middle of the span for a uniformly dis- 
tributed load upon a beam of constant depth. Evidently this 
change of force can only occur by the flange transferring force 

to the web, as was done in the 
g _ b J 

j'^^^T^^ case of the plate girder through 

yJF ^v the flange riveting. AB repre- 

/ \ sents the maximum flange force. 

!. \ Passing from AB to EF the flange 

A _ 1 force has been reduced by the 

I • amount GF which has been trans- 

IG ' 234 ' f erred to the web. In the case 

of reinforcing steel the connection between the steel and the 
concrete must be strong enough to carry the force GF. This 
may be accomplished either by the adhesion between the con- 
crete and smooth or plain bars, or corrugated or twisted bars 
may be used to increase the strength of the bonding. 
Considering plain round bars and letting 

u = unit bonding stress, pounds per square inch, 
8 = diameter of the bar, inches, 
/ = length of bar, inches, 

then t ' 8 *u X- = — Xf 8 , from which - = **— 

24 8 2U 

This means that to develop the full tensile strength of a round 

bar its total length must be - J * J — times the diameter of the bar. 

2 • u 

This relation applies also to square bars. Illustrating this with 



REINFORCED CONCRETE 221 

the f-in. square bars used in a former example, and allowing 
60 lbs. per sq. in. for the working bond stress, we have 

/ / s , 0.875 X 16,000 , . 

- = - jL2 — or / = — " — i — = 116. 5 ins. 

8 2 • u 2 X 60 

This means that under the given conditions to develop the full 
strength of the bar, it must have a total length of at least 116.5 
ins. and must be placed to have a length of not less than 58J ins. 
on each side of the center of the span. A bar's length beyond 
the point at which its allowed tensile stress is developed must be 

The bond stress varies along the bars; in fact, the increment 
of change is that of the flange force, and, similarly to the change 
in horizontal shear, we have 

VXi = (F s -F 2 )j-d, .\ F,-F 2 = -X- y 

J -d 



The unit bonding stress = 



F z -F 2 



u = 



surface of bars for unit length 
V 



j -d (2 circumference of bars) 

The bonding strength of bars may be reinforced by the use 
of corrugated or twisted bars and by hooks made at the end of 
the bars. These hooks when semi-circular and bent around a 
diameter of from 3 to 5 times the diameter of the bar, if properly 
imbedded in the concrete, will resist a pull beyond the elastic 
limit of the bar before the hook loses its grip. Right angled 
bends, when imbedded with sufficient concrete behind them to 
prevent kicking back, are also very effective. 

Lengths of Reinforcing Rods 

The lengths of the reinforcing bars besides being limited by 
the bond stress, as just explained, will also vary in length for 
the same reason the flange plates of girders do. That is, as the 
flange force decreases from the center towards the supports the 



222 



GRAPHICS AND STRUCTURAL DESIGN 



reinforcing area can be reduced. In Fig. 235, CD is the span 
and AB represents the total reinforcing area, A, at the center of 
the span. If 4 bars are used, the line AB is divided into 4 
equal parts, each division representing the area of one bar; the 
lengths of the several rectangles, determined by the intersections 
of the horizontal lines with the parabola CBD* give the respec- 
tive lengths of the bars. In practice the bars instead of being 
cut off are frequently bent up, run to the top and then to the 
end of the beam, as in Fig. 236, thus reinforcing the upper flange 
of the beam over the support, where, owing to the monolithic 
character of a concrete beam, there is restraint and consequent 



, 


>< Length H 

B 




1 — 


1 — 






/"Bar *1 ^X. 




/ •< *= 2 


., 


/ 


" #3 




1 


/ . * 


." ?4 . . 


z\ a 


rN 




i/1 


K=r- 


A 
-Span 

Fig. 235. 


4 D 


% — 


Span 

Fig. 236. 


_ 4 



tension. The bars are usually bent up in pairs. The area run 
over the supports generally exceeds 25 per cent of the gross 
area A ; the remainder of the bars are continued through the 
lower flange and over the supports. If it is preferred to calcu- 
late the length of the bars the following formula may be used, 



wi 



where 



I = length of bar in lower flange between bends, inches. 

a = area in square inches of bars, including all shorter 
bars and the bar whose length is desired. In 
Fig- 235, if the length of bar No. 3 is desired, 
a =area of bars Nos. 1, 2 and 3. 

* Note. — Using a parabola assumes a uniform load on a supported beam. It 
may be replaced by the curve of maximum moments for other than uniform loads. 



REINFORCED CONCRETE 223 

A = area in square inches of the total reinforcing at the 

center of the span. 
L = span of beam, inches. 

WL 

When the moment is g per cent of— —the correction may 

8 

be made by multiplying the value of / found above by Vg. 

The reinforcing bars must be spaced to have sufficient con- 
crete around them to carry the web forces. This spacing must 
also be ample to permit of the thorough compacting of the con- 
crete between and around the surfaces of the bars. The clear- 
ance between the surfaces of adjacent bars should generally be 
not under 1 inch nor if diameters of the bar. 

Where the fireproofmg required by the Specifications, para- 
graph 207, does not have to be enforced, the following clearances 
between the nearest surface of the rods and the surface of the 
beam may be followed. 

Depth of slab Clearance between 

or beam, rods and nearest 

inches. beam surface, 

inches. 

Under \\ f 

5 to 8£ 1 

9 to 12 \\ 

13 to 18 \\ 

18 to 20 if 

Above 20 2 

Design of a T Beam 

A T beam spans 20 ft. o ins.; the beams are spaced 10 ft. 6 ins. 
and carry a uniform dead load of 30,000 lbs., while the uniform 
live load is 75,000 lbs. 

The slab thickness is 7.5 ins., d = 29 ins., and the width of the 
stem h' is 20 ins. Assume the width of slab acting as flange at 
80 ins.* The reinforcing bars are four i|-in. round bars and five 
if -in. round bars. 

The total area of the steel is 

Four i|-in. round bars = 4 X 0.99 = 3.96 sq. ins. 
Five i|-in. round bars = 5 X 1.23 = 6.15 sq. ins. 

Total 10. 1 1 sq. ins. 
* Specification § 213 would have limited this to 60 ins. if enforced. 



224 GRAPHICS AND STRUCTURAL DESIGN 

These bars to be spaced not less than 2§ times their di- 
ameters. 

From the dimensions given we have 

a 29 

10.11 

P = ^ = 0.0044. 

r 80 X 29 



From the curves we find that k = 0.305 and that j = 0.90, 
also -J? = 34. 

Wl 
The bending moment on the beam being assumed as M = — > 



,, 105,000 X 20 X 12 . ,, 

M = — — = 3,150,000 in. lbs. 

8 

M = A*f s • ; • d: f a = ' = 12,000 lbs. per sq. in. 

J J ,J 10.11 X0.90 X 29 ' * H 

Bent Bars. — Assuming that four i^-in. round bars are placed 
above the five ij-in. round bars and that these four bars are 
bent alike, the length of the four bars between the bends will be 



^.i V /|= 24 o V /^ = IS oins. 

Had it been desired to bend up bars 1 and 2 before bars 3 and 4 
were bent their lengths I would have been 



li = L\ — = 240 1/ = 106 ins. 

V A ▼ 10.11 

The minimum length of the bars to develop the proper bonding 
strength is given by the formula 

j 8 Xf a 1. 125 X 1 2,000 

I = ^ = = 00 ins. 

2 X u 2 X 75 

It is seen that the 106 ins. required for flange strength provides 
ample bonding strength. 



REINFORCED CONCRETE 



225 



Web Reinforcement 

The maximum shear at the center of the span will occur when 
the beam is covered with the live load from its center to one of 
the supports. 

The load on one-half of the span then is 37,500 lbs. The live- 
load reaction (Fig. 237) is 

Ri . aMooj^ = 937s lbs _ 

20 
The maximum shear diagram is then given by Fig. 238. 

Fig. 237. 




The straight-line approximation, ab, is commonly used. 

Figure 239 represents a reinforced-concrete T beam from the 
center of the span to the right support. 

From the point i, the intersection of the center line of the span 
and the neutral axis of the beam section, draw ij, making an 
angle of 45 degrees with the horizontal axis of the beam. The 
unit shear at any point below the axis equals the unit diagonal 
tension; at the center of the span this is 



9375 



9375 = . 
b' Xjd 20 X 29 X 0.9 



18 lbs. per sq. in. 



226 



GRAPHICS AND STRUCTURAL DESIGN 



A line ih is laid off from i at right angles to ij and to a length 
representing 18 lbs. Through the point k, at the junction of the 
neutral axis and the line of reaction at the right support, draw jk 
at right angles to ij. The unit end shear is 



52,500 = 52,500 
b' X jd 20 X 29 X 0.9 



100 lbs. 



This value is laid off to scale in jb. hb may be joined by a 
straight line or the true curve may be used as explained in Fig. 
238. The former is done here, it being as accurate as the prob- 
lem warrants. If it is assumed that the concrete can carry 



C.L. 



T-Beam. 




ffffff 



4'f \>T\\ 



be 



3 112 i 

• • • • 



d 




DIAGRAM OF MAXIMUM SHEARS. 

Fig. 239. 

30 lbs. per sq. in. lay oftja to scale representing 30 lbs. and draw 
the line ag parallel to ij. The bars are bent after their lengths 
in the lower flange exceed the distances previously calculated as 
necessary, one-half these lengths being used as the lengths are 
measured from the center line C.L. The area of the trapezoid 
abdc multiplied by b' gives the load carried by the rods 3 and 4. 
The mean ordinate of the trapezoid is 

cd -f ab _ 52 + 70 
2 2 



= 01. 



ac 



19 ins. The area abdc is 61 X 19 = 1160 lbs. 



REINFORCED CONCRETE 227 

The total load on the bars 3 and 4 = 1160 X &' = 1160X 
20 = 23,200 lbs. Hence the unit tension in the bars is 

2 "2 200 

— — = 11,700 lbs. per sq. in. 

2X0.99 " F H 

The tension in the bars 1 and 2 will be considerably less, but 
this is made necessary, as the spacing must be less than d, or 
29 ins., and the length of these bars in the lower flange measured 
from the center line of the span must exceed 53 ins. The bent 
rods are ordinarily assumed as acting approximately through the 
center of gravity of the trapezoid representing the load they 
carry; thus bars 3 and 4 should pass approximately the center of 
gravity of the trapezoid abdc. 

Although the discussion errs on the side of safety it should be 
remembered that, the diagram being one of maximum shears, the 
shears given are not in existence at the same time. There is still 
a diagonal tension ef'g not cared for; this will be provided for 
with vertical stirrups and the whole beam will be strengthened 
by their insertion across the span. The vertical stirrups are 
also useful in holding the horizontal bars in place. 

Had the web-stress been assumed as being re- 
sisted by the vertical stirrups and these stirrups 
taken as J-in. round bars bent as shown in Fig. 
240, making four sections of the rod that would 
have to be broken, and allowing 16,000 lbs. per 
sq. in. as the permissible fiber stress the stirrup IG ' 24 °* 
value would be T — 4 X 0.196 X 16,000 = 12,540 lbs. The 
vertical shears are given by the diagram of maximum shear 
(Fig. 238), and have the following values at the several distances 
from the center of the span. 

Distance from center Max. shear, 

of span, feet. pounds. 

8 43,875 

6 35,250 

4 27,000 

2 18,000 




228 GRAPHICS AND STRUCTURAL DESIGN 

The minimum spacing will be at the supports and will be 

gsB 0.875 d X T = 0.875 X 29 X 12,540 _ g ing 
V-o.&7sXicXb'xd 52,500- (0.875X50X20X29) ~ 

The spacing 8 ft. from the center of the span is approx. 11 ins. 
The spacing 6 ft. from the center of the span is approx. 16 ins. 
The spacing will be made, starting at the piers and running 
towards the center, 2 spaces at 8 ins., 2 spaces at 10 ins., and 
from there to the center the spacing will be 12 ins. This makes 

all the spacing less than-, or— = 14.5 ins., which would be the 

2 2 

maximum for the best practice. The vertical stirrups might 

have been made much lighter considering what the bent rods do 

to strengthen the web, but in this case the vertical stirrups have 

been figured as resisting the entire web tension, the bent rods 

will strengthen the web and will strengthen the beam to resist 

bending by resisting a negative moment over the supports. 

Economic Depth op Concrete Beams 

The best practical depth to make a concrete beam, T beam or 
girder will depend upon the relative cost of the concrete and steel, 
upon the head-room or clearances, upon the necessary width to 
properly enclose the reinforcing bars and upon the required web 
area to care for the web stresses. Ordinarily where the beam is 
of sufficient importance the comparison of several beam sections 
will readily indicate the most satisfactory proportions. Tur- 
neaure and Maurer suggest the following formula for estimating 
the economic depth of T beams: 



H 



r XM 



f.xv 

Here r = ratio of the cost of equal volumes of steel and concrete. 

The remaining symbols have the same significance as given 
on page 197. 

Assume several different widths of stem and estimate the cor- 
responding economic depths by the formula. The ratio r will 



REINFORCED CONCRETE 229 

vary with different places and will ordinarily range from 50 to 
80. Where the ratio is uncertain the mean of the above values 
will probably be a fair assumption. 

Deflection of Concrete Beams 

Owing to the varying moments of inertia along the span of a 
concrete beam together with the fact that the modulus of 
elasticity varies with the unit stress the calculation of the de- 
flection of a concrete beam is at the best but an approximation. 
Mr. G. A. Maney in the Transactions of the American Society 
for Testing Materials, July, 19 14, suggests the following simple 
formula whose accuracy, he states, has been checked by numer- 
ous tests on beams : 

f=k-(e e + e 8 ). 
a 

Here / = deflection. 

k = a constant varying with the loading and nature of the 

support. 

/ = length of span. 

d = depth of beam, to center of steel. 

e c = unit-deformation in extreme fibers of concrete,^- 

e 8 = unit-deformation in extreme fibers of steel, -~f • 

The constant k for simply supported beams has the following 
values : 

Uniform loading k = 0.1041; central loading k = 0.0833; third 
point loading k = 0.1065. In the case of fixed beams, for uni- 
form loading k = 0.0313; central loading k = 0.0416 and third 
point loading k = 0.0347. 

Columns 

Concrete columns are reinforced in one or both of two ways: 
(1) by steel rods paralleling the axis of the column and (2) by 
spirally wound metal bands. In the first case the metal shares 



230 GRAPHICS AND STRUCTURAL DESIGN 

the load with the concrete, while in the second case the bands 
strengthen the concrete by preventing lateral expansion which 
is assumed as occurring when the column is shortened along its 
axis by the load. As it is customary to tie the longitudinal 
reinforcements in place with steel wires or bands it virtually 
amounts to combining both forms of reinforcements. 

The following is the usual discussion of the strength of columns 
having longitudinal reinforcements. 

The nomenclature is 

P = total load on column, pounds. 
A c = area of concrete inside reinforcements, not including steel, 

square inches. 
A 8 = area of steel, square inches. 

A =A g + A c . 

.. A 8 
p = ratio, -— • 

Jx 

If the column with longitudinal reinforcement is loaded, its 
length will be altered, the steel and concrete being shortened the 
same amount. The unit reduction is 

' A = ^ = V 
E, E c 

from which /,=/,Xtt = «X/.. 

■tiic 

The total load carried by the column then is 

P = A 8 ./. + Ac 'fa = A s . nf c + Ac -f c , 
and P = (A s • n + A c )f c ; also P = A -f c [1 + p (n - 1)]. 

The working dimensions of the columns are generally assumed 
as those inclosed by the longitudinal or band reinforcements. 

The proportions of concrete columns ordinarily work out so 
large that their ratio of length to smaller dimensions will usually 
be less than 15, thus making them short columns. The design 
may be safely made upon the basis just given, even where length 
divided by the smaller side reaches 25. 



REINFORCED CONCRETE 



231 




Fig. 241. 




Fig. 242. 




Fig. 243. 



Fig. 244. 



m 



Fig. 245. 



Fig. 246. 



g> 



CROSS SECTION OF BAR 



Fig. 247. 



232 GRAPHICS AND STRUCTURAL DESIGN 

The usual coating of concrete is commonly put on the column 
outside the reinforcing steel to protect it and act as fireproofing. 

A few of the numerous forms of metal reinforcements for con- 
crete are shown in Figs. 241 to 247 inclusive. Fig. 241 shows a 
small section of expanded metal; this is made in varying sizes; 
the smaller, called metal lath, is intended to replace wooden 
laths. For concrete reinforcement the 3-in. mesh expanded metal 
is better. It is made in weights having a cross-sectional area of 
from 0.06 to 0.60 sq. in. per ft. of width. The triangular-mesh 
steel- wire reinforcement illustrated in Fig. 242 serves about the 
same purposes as the expanded metal. These reinforcements 
have the advantage of reinforcing in both directions, that is, 
across the span and at right angles to it. 

Figure 243 is the Ransome twisted bar, Fig. 244 is the Thatcher 
bar and Figs. 245 and 246 are corrugated bars. These bars are 
designed to insure proper bonding between the steel and the 
concrete. 

Figure 247 is the Kahn trussed bar; flanges along the side of 
the bar are bent up to resist the stresses in the web of the beam. 

The Pittsburg Steel Products Company make a reinforcing 
frame consisting of steel bars through the lower part of the beam. 
To these bars on each side of the middle of the span are welded 
shear bars bent up and toward the adjacent support at an angle 
of 45 degrees. 

To the upper end of these shear bars are welded horizontal 
bars which run back over the support. These frames are made 
in standard sizes and the reinforcement in the upper flange over 
the supports is said in all cases to be at least 25 per cent of the 
metal in the lower flange. These frames are built up for certain 
spans and are placed completed in the forms about which the 
concrete is then poured. 



CHAPTER XV 
FOUNDATIONS 

The designer recognizes two kinds of foundations, those for 
structures and others for machines. The function of the former 
is to distribute the pressure properly upon the soil and where 
settlement is inevitable to design the foundations that, as far 
as possible, such settlement shall be uniform throughout the 
structure. It may also maintain the structure upright against 
wind or other forces tending to upset it, as in the case of a chim- 
ney. In machine foundations besides these functions the mass 
of the foundation may play an important part in absorbing 
shock or may even prevent the motion of a machine when acting 
on other machines upon foundations external to its foundation. 
Properly balanced rotary converters and turbo-generators re- 
quire only rigid support, while belt-driven machines, engines 
driving rolling mills, mill housings, etc., must be anchored. 

With modern concrete floors, foundations for most machine 
tools may be dispensed with, the machines being placed where 
desired and rag bolts used to hold the machine to the 
floor. 

The mass of a foundation to limit the vibration of an engine or 
other rapidly moving and imperfectly balanced machine within 
definite limits can only be determined when complete informa- 
tion of the design of the engine or machine is available. The 
determination of this mass therefore is properly the work of the 
engine or machine designer and not within the scope of this book. 
Numerous rules of thumb have been used in determining the 
minimum weights of foundations. One of these is to make the 
foundation weigh 1.5 times as much as the weight of the engine 

2 33 



234 GRAPHICS AND STRUCTURAL DESIGN 

or machine. Another rule credited to E. W. Roberts is that 
F = 0.21 EVN, where 

F - weight of the foundation, in pounds. 
E = weight of the engine, in pounds. 
N = R.P.M. 

The wide variations in these two rules show how unsatisfactory 
such formulae are. Vertical engines require heavier foundations 
than horizontal engines of the same capacity and speed. 

The foundations of machines subjected to much shock should 
be kept free from other foundations and the building. 

Although considering only pressure and settlement the best 
foundation bed is rock, yet vibrating machines placed directly 
upon the rock may result in the transfer of the vibrations through 
the rock for considerable distance. The usual precaution under 
such circumstances is to place 2 or 3 feet of sand held in a pocket 
in the rock; this may be either a natural pocket or it may be 
made of concrete. The foundation is then built upon this cushion 
of sand. 

In setting an engine or important machine the foundation is 
brought to within f in. to i| ins. of its final top. The machine 
is then lined up by driving metal wedges between the foundation 
and machine bed. When the machine is properly in line a dam 
is built around the foundation top and a thin grout poured in, 
filling up the space between the bed plate and the foundation. 
After the grout hardens it firmly secures the bed and the founda- 
tion regardless of the finish of the engine bed or foundation top. 
When the grout is properly set the wedges may be removed 
and the spaces left by them filled up with cement mortar. The 
foundation bolts are preferably located and held in place during 
the construction of the foundation by templates of the bed plate. 
The bolts may either be set right in the concrete or may be 
placed in pipes whose inside diameters somewhat exceed the bolt 
diameters and the pipes are then set in the concrete. The bolts 
are carried by the templates; the space between the top of the 



FOUNDATIONS 



235 



pipe and the bolt is filled with burlap, waste or paper, to prevent 
the concrete filling the clearance space between the pipe and the 
bolt. 

This clearance permits some adjustment of the bolts should 
the bed plate disagree with the template. The foundation bolts 
should be run well down into the foundation. Frequently the 
nuts or keys are placed in pockets made accessible from the out- 
side of the foundation. This permits of the ready replacement 
of a bolt should it be found to be necessary. 

In some cases what is known as a rust joint is .made between 
the foundation and the bed plate. This is made by rusting cast- 
iron chips together with sal-ammoniac. Melted sulphur was 



u 





BOLT 

Fig. 248. 



KEY 

Fig. 249. 



WASHER 

Fig. 250. 



also much used before cement grout became so common. The 
foundation bolts generally have the end above the foundation 
threaded for a nut; the other end may either have a thread and 
nut or be slotted for a key. The washers in the foundation are 
of a great variety including scrap channels, angles or rails. 

Cast-iron washers may also be used but of whatever kind they 
should be large to distribute the pressure well over the concrete. 
Fig. 248 illustrates a slotted bolt, while Fig. 249 shows the usual 
key and Fig. 250 the cast-iron washer. In other cases the ad- 
justment of the bolt is provided for only at the top of the bolt 
for a distance of a couple of feet. This is done by using a short 
piece of pipe similar to the way mentioned before or by setting 
a wooden casing around the bolt which leaves a pocket in the 
finished foundation around the top of the bolt. The foundation 



236 



GRAPHICS AND STRUCTURAL DESIGN 



should be carried to proper soil and always below the frost line. 
The load upon the soil should be calculated to see that it does 
not exceed that permissible. The allowable pressure upon the 
soil is commonly given by the building codes where the founda- 
tion is within a city's limits. 

It may frequently be had by obtaining data about neighboring 
foundations. The following table from " Baker's Masonry " is 
also available as a general guide. 



Soil. 


Bearing power, 
tons per sq. ft. 


Soil. 


Bearing 
power, tons 
per sq. ft. 


Rock, thick layers 

Rock, equal to best brick. 
Clay, thick beds, dry. . . . 
Clay, thick beds, moder- 
ately dry 


200 or over 

15-20 

4-6 

2-4 
1-2 


Gravel and coarse sand. . 
Sand, compact 


8-IO 
4-6 
2-4 
i-I 


Sand, clean 


Alluvial soils 






Clay, soft 









Where the engine or machine is an important one and the soil 
of doubtful value, piles are frequently resorted to. The subject 
of piles is considered under building foundations. 



4 



ii:' 



1 11 1 
iiiii 

W 





Hole 




Fig. 251. 



Figure 251 illustrates a foundation for a machine. The bolts 
have been set in pipes and run through to pockets accessible 
from the outside so that the bolts could be replaced if necessary. 
This foundation shows the base extended to reduce the pressure 
upon the soil and a semicircular hole has been cut through the 
foundation to reduce its weight. 



FOUNDATIONS 



237 



In another class of foundations for machines the foundation 
requires weight to prevent the overturning of the machine. 
Fig. 252 illustrates such a foundation for a crane. The load of 
33,000 lbs. is to be carried at a radius of 33 ft. The frame and 




1.45 



Distribution of 
Pressure on Soil 



Fig. 252. 



machinery is assumed as weighing 11,000 lbs. and as acting at 
a distance of 6| ft. from the post center. The foundation is 
circular and has been assumed as weighing 425,000 lbs. The 
total weight on the soil then will be the sum of these three 
amounts, 33,000 + 11,000 + 425,000 = 469,000 lbs. 



238 GRAPHICS AND STRUCTURAL DESIGN 

The distance of the line of action of the resultant of these forces 
can be found by taking moments about the vertical axis of the 
post; hence 

x = (33 X 33,000) + (6.5 X 11,000) = 2 f t 
469,000 

The maximum pressure should now be determined as is done 
for a chimney (see page 254). The foundation has been assumed 

as circular. The kern radius of a circle is — • Here then r = — 

8 8 

= 2.0 ft. The resultant, therefore, falls outside the kern. Re- 
ferring to Fig. 270, we first find 

then • a = 0.355 an d W = ap e D 2 y 

W 469,000 „ 

or p e = —— = y \ = 5160 lbs. 

aD 2 0.355 X 16 2 

Whether or not this maximum pressure is permissible will de- 
pend upon the character of the soil or nature of the foundation 
under the section considered. The curves give <f> as 41 per cent 
and the neutral axis therefore falls 16 X 0.41 = 6.55 ft. to the 
left of the vertical axis, or 1.45 ft. to the right of the left edge 
of the foundation when the full load is in the position shown in 
the figure. The distribution of this pressure is indicated below 
the assumed foundation bottom. 



Building Foundations 

Since settlement of foundations is bound to occur, unless 
foundations rest on bed rock, the foundations of a structure 
should be designed as far as possible to lead to uniform settle- 
ment. The effect of unequal settlement is frequently evident 
where chimney foundations have been carried into wall footings. 



FOUNDATIONS 239 

Here the settlement of the chimney may cause considerable 
cracking of the wall. 

In designing foundations it is considered better to design the 
areas proportionally to the dead loads, which act continuously, 
and then see that they are sufficiently large to carry the combined 
dead and live loads without bringing excessive pressure upon the 
soil. The dead loads acting continuously will exert a greater 
effect upon the settlement of the foundation than live loads 
which may act but infrequently and then possibly be only a 
small part of the assumed live loads. 

When the footings are proportioned for both dead and live 
loads only such portion of the live loads should be considered 
as may be assumed as acting continuously. It is also of prime 
importance that as far as possible the center of pressure on the 
soil shall coincide with the center of gravity of the footing. The 
building codes of the various cities give the requirements of foot- 
ings and these are commonly required to be a certain enlarge- 
ment of the wall upon them. Where these codes are not available 
the footings can be designed using the tables of allowable pres- 
sures previously given. The areas should be enlarged under 
pilasters carrying heavy concentrations. 

Foundations for building columns, in addition to resisting 
vertical loads, may carry the horizontal reactions of wind loads, 
as in the case of the bent for the steel-mill building designed on 
page 121. Here there will be two cases, one (Fig. 253) where 
the column is assumed as hinged and then the horizontal reac- 
tion acts at the column base; the other (Fig. 254) where the 
column is considered as fixed and the point of application of the 
horizontal wind reaction is taken midway between the foot of 
the knee brace and the column base. 

The base is commonly a rectangle. Were there no horizontal 
force the pressure upon the foundation would be uniformly dis- 
tributed. The wind pressure tends to increase the pressure upon 
the leeward side of the foundation and decrease it upon the 
windward side. 



240 



GRAPHICS AND STRUCTURAL DESIGN 



As long as the resultant pressure R cuts the bottom of the 

foundation at a distance not exceeding — from the center of the 

6 

base there will be compression over the entire base. R is the 

resultant of W and H, where W = load carried by the column 

or wall plus the weight of the foundation and E = horizontal 

wind reaction on the column. Since neither masonry nor the 

surfaces of the foundation and soil in contact can be in tension. 





Fig. 253. 

having R pass beyond the middle third will reduce the area of 
the foundation under pressure and will consequently increase 
the pressure on the leeward side, see Fig. 254. 

When R falls in the middle third the extreme pressures on the 
edges of the foundation are given by 

W 

here b = width of the base at right angles to B. 



FOUNDATIONS 



241 



When R falls beyond the middle third p e = 



2-W 



3-b >c 

The distance x may be found graphically, as shown, or by 
moments, since 

H-h 



W • x = H • h or x 



W 



h = distance from bottom of foundation to force H. 

In the second case (Fig. 254) the attachment of the columns 
to the bases must be sufficiently strong to develop the bending 

moment H X - at this point. 

2 

The more general discussion of the kern of a section and the 
distribution of pressure upon the soil is given under chimney 
foundations, page 247. Where 
the base of a foundation ex- 
tends beyond the main shaft, 
Fig. 255, the portion extended 
should be calculated as a can- 
tilevered beam subjected to a 
uniform load p e . This is com- 
monly stated as tons or pounds 
per square foot. This exten- 
sion e is given by e = 4 d\ ^- . 

T Pe 

p = 30 to 50 lbs. per sq. in. in concrete. 

p e = maximum load on soil in pounds per square foot. 

Figure 256 gives a foundation standard adopted by one 
company. Capstone — 1 Portland cement, 2 sand, 4 blast fur- 
nace slag. Foundation — 1 Portland cement, 2 sand, 6 slag, 
minimum bolt diameter — if in. — pressure on metal at base 
of column not over 10 tons per sq. ft. Pressure on capstone 
not over 5 tons per sq. ft. Bearing pressure per square foot on 
gravel 2.5 tons. Bearing pressure per square foot on other soil 
2 tons. 




Fig. 255. 
All dimensions in inches. 



242 



GRAPHICS AND STRUCTURAL DESIGN 




^3 






^-' 



"B" determined by 

Fig. 256. 



I 
Fig. 257. 



Where a shallow and light foundation is sufficient a grillage 
or reinforced-concrete footing can be used. The beam or rails 

















C? 


A 


\ 






/ 




i 


\ 






/ 






b 


\ 






/' 










\ 


k^ 








/ B ' 










/ 

/ 






\ 






/ 
/ 






\ 

\ 










i > 




*"" " 




L5 ^ 



£ 

T 



used in Fig. 257 may be cal- 
culated by assuming the bend- 
ing moment on a tier of beams 
P-e 



as M = 



where P = the 



total load on the column, 
e = the extension of . the 
beams beyond the superim- 
posed beams, or column base. 
Owing to the uncertainty 
about the loading and deflec- 
tion of the beams and the 
column base the load is as- 
sumed as uniformly dis- 
tributed by the column base, 
or upper tier of beams on the 
beams below. 
FlG * 258# In the reinforced-concrete 

footing (Fig. 258) the reinforcement can be only roughly esti- 
mated. One method is to consider the load upon the trapezoid 
abed and assume it as carried by the width be. The load W on 




FOUNDATIONS 243 

jA; the distance x to the center of 

gravity of the trapezoid is x = * X — • 

& +i>i 3 

The bending moment is M = W (A — x). In the formula 
for W, all dimensions are in feet, p e = lbs. per sq. ft. and W is 
in pounds. In the formula for x all dimensions are in inches. 
In the equation for M all the terms are in pounds and inches. 
Having found the moment M the design is readily completed as 
explained under reinforced concrete and as illustrated by the 
design under a reinforced-concrete chimney. 

Under the date of March 31, 1913, Bulletin No. 67 of the 
University of Illinois, entitled " Reinforced Concrete Footings 
and Column Footings," by Arthur N. Talbot, gives the results 
of tests and makes recommendations for the proper designing of 
footings based upon them. For a square footing and column 
base with two way reinforcement the suggestions are as follows. 
The moment on the section at be is assumed as in the preceding 
case (Fig. 258). The resisting moment of the concrete footing is 
calculated considering the area of the steel in a distance across 
the footing equal to the width of the pier be plus twice the thick- 
ness of the footing plus one-half the remaining distance on each 
side of the edge of the footing. This method may also be used 
when the spacing through the middle of the width of the footing 
is closer, or even when the bars are concentrated in the central 
portion. 

With two way reinforcement evenly spaced over the footing, 
it seems that the tensile stress is approximately the same for bars 
lying within a space somewhat greater than the width of the 
pier and that there is also considerable stress in the bars lying 
near the edges of the footing. 

Bond stresses are most important and the bulletin recommends 
that the bars shall not exceed f inch. 

The maximum bond stress will be that given on page 221, 
where V = load abed and the number of rods is that stated above 



244 GRAPHICS AND STRUCTURAL DESIGN 

and assumed as resisting the bending. Hooking the ends of the 
bars increases the bond resistance but is troublesome. Using 
long bars bent into horizontal loops covering the entire footing 
gave high bond resistance. 

Diagonal Tension. — The vertical shear is taken as the 
measure of the diagonal tension. The unit shear is taken as that 
on the sides of a square concentric with that of the footing and 
the column base. The total shear, V, equals the load on the 
footing outside of the square being tested. If D is the depth 
of the footing and A a side of the pier, the square giving the 
maximum shear will have a side (A + 2 D) or its perimeter will 
be 4 (A + 2 D). The critical vertical shearing stress is 

= V 

4 (A+2D)XjD' 

If B is the side of any other square, then 



Vi = 



AB-jD 



The section under the pier will be subjected to punching shear; 
see Specifications, paragraph 2i8d. 

Piles 

Where the sustaining power of the soil is inadequate any of 
the preceding foundations may be placed upon piles. Piles were 
until a few years ago entirely of timber, preferably white oak, 
with a point diameter of at least 6 ins. but frequently specified 
as 8 to 10 ins. Their lengths will vary to suit the particular 
location. During the last few years extensive use has been 
made of concrete piles. Timber piles should be sharpened at the 
point, and are frequently shod with an iron band or point to 
protect them. The larger end is cut off square and is also some- 
times protected with iron caps, hoops or bands. The brooming 
of the butt of a pile interferes materially with driving it, so that 
such piles should be trimmed during the driving to facilitate this 



FOUNDATIONS 



245 



operation. The final or test blows should not be made on a 
broomed pile. 

Piles may carry their load through friction between their sides 
and the surrounding soil, or they may be driven to rock when 
the load may be carried largely by the rock. A pile may there- 
fore fail through crushing of the timber, or, where a pile is driven 
to rock or its equivalent through loose soil, it may fail as a 
column. Specifications limit the load on a pile to from 40,000 
to 50,000 lbs., or 600 lbs. per sq. in. of its mean section. When 
a possible failure as a column is considered the load should be 
calculated for a column whose maximum stress does not exceed 
600 lbs. per sq. in. 



4 



^o;^ concrete ^ ^ .^.^ 




Fig. 259. 



Timber piles when completely submerged in water and not 
exposed to the toredo (shipworm) are apparently preserved 
indefinitely. It is therefore necessary to cut timber piles below 
extreme low water. A bed is then built upon the tops of the 
piles to carry the upper part of the foundation (see Fig. 259). 
The usual spacing of the piles will vary from two to four feet and 
upon the tops of the piles which have been sawed off to a common 
level is placed a grillage of timbers secured to the piles with 
metal pins. The masonry is then placed upon this timber. All 
timber must be below water. Another and later practice, shown 
in Fig. 260, is to incase the tops of the piles in a bed of concrete. 
As in the first case the piles are cut off below the water level. 
The soil around the piles is excavated for two feet and a bed 



246 GRAPHICS AND STRUCTURAL DESIGN 

of sand, say, 12 ins. thick, is put in, and upon this is laid the 
concrete whose thickness depends upon the load to be carried. 
This last method firmly incases the pile tops and distributes a 
part of the load upon the soil. 

Concrete piles have an advantage over timber piles in not 
being attacked by the toredo and in not having to be continu- 
ously submerged. 

There are several types^ of patented concrete piles. The 
Simplex is made in place by driving a hollow tube with a pointed 
end the entire depth required and then filling the hole with con- 
crete as the tube is withdrawn. In the Raymond pile a thin 
tube is driven from which a collapsible case is then removed and 
replaced with concrete. Other forms of reinforced piles are 
made on the ground and when properly aged driven with a pile 
driver. 

There is no very satisfactory method of determining the bear- 
ing power of a pile. The best known, where the pile is not driven 
to refusal, is the Engineering News Pile Formula. This formula 

2 • W • h . 2 • W * h 

is P = for drop-weight hammers and P = — ■ — r— for 

5 + 1 F 6 s+iV 

steam hammers. 

P = safe load in tons, 2000 lbs. 
W = weight of hammer in tons, 2000 lbs. 

h = drop of hammer in feet. 

s = penetration in inches due to the last blow. 



CHAPTER XVI 



CHIMNEYS 

In the design of masonry chimneys it is usual to assume that 

the masonry cannot resist tensile stresses. If the weight of the 

column above the section i-i, Fig. 261 (a), 

is W pounds this load will be uniformly 

distributed on the section when no other 

forces act on the column above this section. 

If then the area is A sq. ins. the uniform 

W 
fiber stress is — lbs. per sq. in. Assuming 

j± 

a force F acting on the right of the column, 
it will tend to reduce the fiber stress on the 
right and increase that on the left of the 
section. The original fiber stress / is com- 
pression. If the force F is increased suffi- 
ciently the fiber stress on the right would 
become zero while that on the left would 
be doubled, as is indicated in Fig. 261 (d). 
The unit stress would vary across the sec- 
tion as shown in Fig. 261 (d), being as- 
sumed as varying uniformly. The stresses 
would still all be compression. Under these 
conditions the resultant R of W and F must 
act through a point a distance q from the 
center of gravity of the section 1-1. Should 
the resultant R act at a greater distance 
from the center of gravity of the section 
than q, which would be due to a still further increase in F, then 
the fiber stress on the right would pass through zero and be- 

247 




Fig. 261. 



248 GRAPHICS AND STRUCTURAL DESIGN 

come tension, should the material resist tension, and the fiber 
stress on the left would be increased but would continue to be 
compression. Should the material not be able to resist tension 
the pressure will adjust itself to the left until the resultant 
stress R coincides with the point of intersection of the resultant 
of W and F with the section 1-1. When the extreme fiber 
stress on the left exceeds the strength of the material, the sec- 
tion 1- 1 will fail. It is usual to assume that the resultant R 
must fall within the distance q from the center of gravity of the 
section. Had the force F been applied at all possible points 
around the column at the same distance y above the section 
1-1 the points of intersection of the resultants R of W and F 
would have described the figure shown black in Fig. 261 (b); 
this is called the kern of the section. If the section 1-1 at no 
point is to be subjected to tension the resultant R must fall 
within the kern for all positions of the force F. 

The moment on the section for a given loading will be the load 
W times the distance from the center of gravity of the section 
to the point of intersection of the resultant R with the plane of 
the section 1-1. 

The allowable distance q from the center of gravity of the 
section to the resultant R that produces zero stress at the extreme 

fibers of the section for that position of F is given by q = > 

A. X 6 

where / = moment of inertia of the section, inches 4 . 

A = area of the section, square inches. 

e = distance from center of gravity of section to extreme 

fibers whose stress is to be limited to zero. 

The kerns for the commonest sections used for chimneys and 
foundations are given in Figs. 262 to 267. 
The minimum radii for the kerns of these several sections are, 

Square, Fig. 262, r = 0.118A. 

b.h 
Rectangle, Fig. 263, r = 6V p + p' 



CHIMNEYS 249 

Triangle, Fig. 267, r 2 = — , and t\ = -- 
12 6 

Octagon, Fig. 264, r min = 0.2256 R (R = radius of corners). 

Circle, Fig. 265, r = - (constant). 
8 

Hollow square, outer side = H, inner side = h. Similar to 
Fig. 262 

rmln = o.n8 #[i + (A) 2 J 

Circular ring, Fig. 266, outer diameter = Z), inner diameter 
= d. 



=![■+©' 



In the case of retaining walls it is necessary to consider the force 
F as acting in one direction; the base is then rectangular and 
the resultant R must not fall farther from the center of the base 

than one-sixth the base width, or if b = width of the base, q = - . 

6 

This is commonly expressed by saying that the resultant pressure 
must pass within the middle third. 

In chimney design in this country the wind pressure is gen- 
erally assumed at 30 to 50 lbs. per sq. ft. upon flat surfaces, and 
at 20 to 30 lbs. per sq. ft. on the projected area of cylindrical 
surfaces. The greater of these pressures considerably exceeds 
any pressure likely to occur in ordinary localities. 

Ordinary masonry chimneys are commonly either square, 
octagonal or round. Square sections are suitable for short and 
unimportant chimneys while by far the greater number of 
chimneys are round. 

Chimney sections are obtained by the use of specially shaped 
brick for the corners of octagonal chimneys and radial brick for 
round chimneys. 

In small chimneys the upper 25 ft. should have a thickness of 
8 to 9 ins. ; this thickness should be increased about 4 to 4 J ins. 




Fig. 262. 




h-^rH 




-1 8'0" 



I* b >\ t 

Fig. 263. 




__2ft 4 




\- 



.Top 



V- 



■f- 



\— 



Ground 



30% 



BRICK CHIMNEY 
Fig. 268. 



(250) 



CHIMNEYS 251 

in each succeeding lower 25 ft. In large chimneys whose diam- 
eters exceed 60 ins., the thickness of the upper 25 ft. should be 
12 to 13 ins. 

Ordinary brick dimensions are, width 4 to 4! ins.; length 
8 to 9 ins. ; thickness 2 to 2 J ins. 

Chimneys are commonly provided with a fireproof lining in 
the lower portion extending up to from one-third to one-half 
the chimney height. This lining, being protected from the wind 
pressure by the main walls of the chimney, can be made quite 
thin, frequently not exceeding from 4 to 8 ins. 

Designing a Chimney 

The method of designing a chimney will be illustrated by 
making the calculations for the section d-d of the chimney in 
Fig. 268. The weight of the material above the section has 
been estimated as 545,000 lbs., being taken at no lbs. per cu. ft. 

The wind pressure has been considered as 25 lbs. per sq. ft. 
of projected area for a round chimney and has been estimated 
as 30,800 lbs. This wind force has been assumed as acting at 
the center of gravity of the trapezoid which is the projected 
area of the chimney above the section. 

The distance from the section d-d to the center of gravity is 

x = b -tlA x d = i5,33 + 20 x 100x12 = . ns 
b + h 3 15.33 + 10 3 

b = base of trapezoid. 

bi = top of trapezoid. 

d = altitude of trapezoid. 

From the intersection of the chimney axis and wind force lay 
off hi representing the weight to scale, 545,000 lbs., and at its 
lower extremity lay off ij to the same scale representing F = 
30,800 lbs. Draw the hypothenuse of this right-angled triangle 
and it will cut the section d-d at I. To insure the masonry at 
this section not being subjected to tension the point / must fall 
within the kern of this section. 



252 



GRAPHICS AND STRUCTURAL DESIGN 



For accuracy it is better to calculate the length Ik than to 
measure it from a drawing on so small a scale. The calculation 
can be made by the use of the similar triangles hlk and hij, hence 

7 , hk X ij 558 X 30,800 

Ik = a = — — -J- = 22 *2_> = 31.5 ins. 

In 545>oco 

M = distance from the section to the resultant wind force. 
The kern radius is given by 

'=fh(l)']=fh®>— 

Since 31.5 ins. is less than 35.6 ins. it is evident that the 
resultant falls within the kern, and the section is subjected to 
compression only. 

The weights have been calculated, using the volume of a 
frustum of a cone as 

h 



3 
h = height of the frustum. 
A = area of the base. 
a = area of the top. 

The following table gives the properties of the several sections. 



Section above. 



At top . 

a-a 

b-b.... 

c-c 

d-d. . . . 
e-e .... 



H 

Top of foundation . 

Base foundation. 



Distance 

from top, 

ft. 


Diameter 
O.D. 
I.D. 


Mean 
diameter. 


jMean 
area, 
sq. ft. 


Weight 
above, lbs. 


Fat 
25 lbs. 


r" 





j io'-o" 
\ 8'-c" 












25' 0" 


| 11' 4" 
( 9' 4" 


10' 8" ) 
8' 8" ) 


28 


77.000 


6,700 


28.6 


So' 0" 


| 12' 8" 
\ 10' 0" 


12' 0" ) 
9' 8"( 


40 


187,000 


13,750 


30.8 


75' 0" 


i 14' 0" 
\ 10' 8" 


13' 4" ) 
10' 4" ) 


56 


341,000 


21,800 


33.2 


100' 0" 


1 15' 4" 
. (ii' 4" 


14' 8" » 
11' 0" ( 


74 


545, 000 


30,800 


35.6 


125' 0" 


{ 16' 8" 
( 12' 0" 


16' 0" ) 
11' 8"( 


94 


803,000 


40,700 


38.0 


150' 0" 


( 18' 0" 
\ 12' 8" 


17' 4" ) 
12' 4 1 


117 


1,127,000 


52,500 


40.1 












fMax.l 


185' 0" 


{ i8 :°::} 

(II' o"J 

f T °p ] 




229 


2,008,650 


67,500 


I 46.8,1 
lMin.f 
I.33.4 J 


percu. ft.. 


1 22' 0", 1 
I Bottom [ 
I 30' 0" J 






2,867,130 


67,500 


j42.5\ 
(Min./ 



12.7 
21.3 
27.3 
31.5 
34.8 
37.9 

33.4 
26.2 



CHIMNEYS 253 

It is seen in the above table that since the resultant pressure al- 
ways falls within the kern the maximum pressure never exceeds 
twice the fiber stress due to the weight above the section assumed 
as uniformly distributed over that section. Considering the sec- 
tion at/-/, the load above that section being 1,127,000 lbs., and 
the area of the section 117 sq. ft., the load per square foot is 

^2^2 = 9630 lbs . 
117 

Twice this is well below the allowable, or 144 X 150 = 21,600 
lbs. per sq. ft. 

Some designers prefer to use the inertia of the section in 
making their calculations. Making the calculations for the 
section 150 ft. from the top we have 

The section modulus of the ring 

= I = < JT (£> 4 ~ d 4 ) = 7T (D 2 ~ d 2 ) (D 2 + d 2 ) 

e 32 D 4X8D 

Since - {D 2 — d 2 ) is the area of the ring section which we can 

4 

call A we have 



e 8\ T Z)/ 



The dimensions for the section 150 ft. from the top are A = 117 
sq. ft., D = 18 ft. o ins., d = 12.67 ft., hence 



I 117/ ' i2.67 2 \ 



The projected area of the chimney is X 150 = 2100 sq. 

ft. The wind pressure on this area is 2100 X 25 = 52,500 lbs. 
The distance x from the section to the center of gravity is 

2fx^=68ft. 

28 7, 



254 GRAPHICS AND STRUCTURAL DESIGN 

The moment of the wind about the section is 52,500 X 68 = 
3,570,000 ft. lbs., since 

- Me 1, 570,000 „ fA 

f = ~r = — = 9 IO ° lbs - P er s q- ft. 
1 393 

The direct pressure upon this section is — — — = 9630 lbs. 

117 

per sq. ft. The maximum pressure upon the leeward side is 

9630 + 9100 = 18,730 lbs., while that on the windward side is 

9630 — 9100 = 530 lbs. per sq. ft., both being in compression. 

Some designers assume an allowable tension on the windward 
side of one-tenth the maximum compression on the leeward side. 
It is then understood that the material does not exert this tensile 
fiber stress but that the effect is to make the resultant com- 
pression pass outside of the kern and increase the maximum unit 
pressure. This can be understood by referring to Figs. 261 (a) 
to 261 (f). 

The determination of the maximum pressure under these 
circumstances presents considerable difficulties. As long as the 
resultant pressure at any section fell within the kern the mar- 
gin of security would be given by the ratio of the ultimate 
crushing strength of the material to the extreme fiber stress 
in compression at that section. Keeping the resultant within 
the kern affords a ready means of determining the maximum 
fiber stress, thus assuring safety. 

The maximum pressure upon the foundation will now be 
determined. The foundation being 30 ft. o ins. square the min- 
imum radius of the kern is 

r = 0.118 X 30 X 12 = 42.5 ins. 

As the resultant passes 26.2 ins. from the center line of the 
foundation it falls in the kern and the maximum unit compression 
will not exceed twice the unit uniform load on the foundation 
when no wind is blowing. 

The maximum unit pressure will be on a corner when the wind 
blows parallel to a diagonal. The moment on the foundation 



CHIMNEYS 



255 




Fig. 269. 




Fig. 270. 



256 



GRAPHICS AND STRUCTURAL DESIGN 



section is the product of the total weight on the soil multiplied 
by q or 

m 2,867,000 X 26.2 

AZ = -* — u = 6,260,000 ft. lbs. 



12 



The resistance of a square referred to a diagonal as an axis is 

- = 0.118 A 3 . 



The base being 30 ft. square and keeping the resistance in feet 3 
we have, 

- = 0.118 X30 3 = 3186. 
e 

e . . M • e 6,260,000 , ,, ,, Al ,. 

Since/ = — — = — — -J- — = 1965 lbs. per sq. ft., the direct 
1 3180 



pressure is 



2,867,000 
900 



3190 lbs. per sq. ft. The maximum 



compression then is 1965 +3190 = 5155 lbs. 
per sq. ft. The compression on the opposite 
corner is 3190 — 1965 = 1225 lbs. per sq. ft. 

These loads will be satisfactory if the soil can 
carry 2J tons per sq. ft. 

When the resultant of the wind pressure and 
the chimney weight passes outside of the kern 
of the bottom of a chimney the maximum pres- 
sure on the corner of a square or the circum- 
j t ference of a circular section may be found by 
using the curves given in Figs. 269 and 270. 

Fig. 269 represents a square base, Fig. 270 a 
circular base. 

* — D = diagonal of square and diameter of cir- 
cle. 
h = side of square. 
q = distance from center of section to resultant of wind 
and weight. 
a- a = neutral axis, axis of zero pressure. 



T 



Fig. 271. 



CHIMNEYS 257 

F = total wind force on chimney, acting at the center of 

gravity of the projected area of the exposed portion of 

the chimney. 

W = total load on the soil, pounds. 

hi =. distance from force F to bottom of the foundation, 

p 
inches. See Fig. 271. # (inches) = hX^r - 

Self-sustaining Steel Chimneys 

A self-sustaining steel chimney, Fig. 272, is held upright by 
the resistance to overturning offered by the chimney, its lining 
and foundation. 

The wind pressure is commonly assumed as 50 lbs. per sq. ft., 
acting upon the side of a square chimney, or 25 to 30 lbs. per 
sq. ft. of vertical projection upon round chimneys. 

Usually the thickness of the upper sheets is determined for 
durability rather than for strength, being made not less than 
T 3 g in. and frequently \ in. thick. This thickness is then in- 
creased by yg in. every 30 or 40 ft. Some designers vary by 
g 1 ^ in. instead of by T *g in. At the lower part of each 30 or 40 ft. 
section the fiber stress induced in it by the bending moment 
due to the wind pressure can be found and the section altered in 
thickness or length if deemed advisable. 

The rivet spacing is made small to assure tightness, being not 
less than 2.5 times the rivet diameter nor more than 16 times 
the thickness of the plate. For most ring sections this gives 
excessive rivet strength. 

An assumed ring section may be checked as follows: 

M = bending moment in inch pounds on the section due to the 

assumed wind pressure above that section. 
D = outside diameter of the chimney in inches. 
h = distance from a ring section to the chimney top, inches. 
t = thickness of shell, inches. 

F — total wind pressure acting upon any portion of the 
chimney, pounds. 



Fig. 272. 




BRACKET 



STEEL CHIMNEY 



<2si) 



CHIMNEYS 259 

The chimney above the given ring section acts as a cantilever 
beam, and the bending moment equals the total wind pressure 
upon the vertical projection of the chimney above this ring 
section multiplied by one-half the distance from the ring section 
to the chimney top. The chimney will be assumed 66 ins. 
I.D. and 100 ft. high. The bell section is 9 ft. high and the 
section to be examined is 91 ft. from the chimney top. The 
total wind pressure on the vertical projection of this part of 
the chimney is F = 5.5 X 91 X 25 = 12,510 lbs. and 

OI ^" T 2 

M = 12,510 X- = 6,830,460 in. lbs. 

2 

The value of - for a ring section is 
e 

I _ it (D* - d*j 
e 32 XD 
where D = outside diameter. 
d = inside diameter. 
Where /, the thickness of the shell, is small compared with D, 
this value may be approximated as 

T -=^Dt(D- 3 t)* 
e 5 

-for the case in hand is - = - X 66.625 X -^ (66.625 — 0.9375) 
e e 5 16 

= 1094. 

, Me 6,8^0,460 , „ 
/ = — - = * * = 6250 lbs. 
/ 1094 

This fiber stress is on the gross section. If the net section at the 

rivet circle is x per cent of the total section then the maximum 

fiber stress between rivets is 

, 6250 -r o r 6250 

L = — —, or, if x = 80 per cent, f m = —-*- 

x .00 

= 7800 lbs. per sq. in. 
As these fiber stresses are satisfactory the iV m - sne ^ will be used. 

* Note. is also frequently taken as the area inside the shell multiplied by 

the thickness of the shell, all measured in inches. 



260 graphics and structural design 

Riveting Ring Seams 

5 = pitch of rivets, inches. The force acting on the rivets 
per inch of circumference is i X / X /, and to this must be added 
the weight of the shell i in. wide from the given section to the 
top. The calculation of this weight can be facilitated by remem- 
bering that a J-in. plate 12 ins. X 12 ins. weighs 10.2 lbs., then 
estimating the weight of a vertical strip 12 ins. wide and finally 
dividing the weight found by 12. 

For the section 91 ft. from the top, 1 X ^ X / = 1 X ^ X 6250 
= 1950 lbs. 

1. 30 ft. of Y6" m - plate, 30 X 1 X (| X 10.2) = 230 

2. 30 ft. of J-in. plate, 30 X 1 X (1 X 10.2) = 306 

3. 30 ft. of fV-in. plate, 30 X 1 X (f X 10.2) = 383 



Total 919 

Adding 10 per cent for rivets, laps, etc. 91 



1010 



or = 84 lbs. per in. The total force per inch of circum- 

12 

ference acting on the rivets is 1950 + 80 = 2030 lbs. Allowing 

10,000 lbs. per sq. in. in single shear and 20,000 lbs. per sq. in. 

in bearing the value of a f-in. rivet in a Ye-in. plate is 4420 lbs., 

and the rivet spacing required is — — =2.18 ins. or, say, not 

2030 

exceeding 2 J ins. 

A double row of staggered rivets will be used at this seam 
spaced about, but not exceeding, 2J ins. Ordinarily the rivet 
spacing need be determined for only the lowest row of rivets in 
sheets of the same thickness. 

Riveting Vertical Seams 

The riveting in the vertical seams must provide for the shear 
along the neutral axis of the chimney, and as the riveted edge 
may be in compression the rivets must be spaced to prevent 



CHIMNEYS 261 

the buckling of the plate between them; this demands that the 

rivet spacing shall not exceed 16 times the thickness of the plate. 

The unit shearing stress at any point along a cylindrical beam 

r - v - r 

e 
where /, = unit shear in pounds per square inch at the section. 
V = shear at right angles to the neutral axis, pounds. 
r = radius at the section, inches. 

- = resistance of the section, inches 3 . 
e 

In this chimney at the section under consideration the unit 
shearing stress is 

- V 'T I2,SlO X^ „ 

/. = —f- = — ^ & = 380 lbs. 

I 1094 

e 

This per inch of chimney height is 380 X j& = I2 ° lbs. 

The rivet spacing for shear then is — 37 ins. 

120 

This spacing of course is not permissible, 16 times the plate 
thickness limiting the spacing to 5 ins. It generally will be the 
buckling of the plate edge that will determine this riveting. 

The ring sheets are preferably of one piece, but in large chim- 
neys this becomes impractical both for manufacture and ship- 
ment. 

The base of the chimney is usually a bell, and is preferably 
a frustum of a cone, although sometimes flared to improve the 
appearance. The former shape is a much simpler and cheaper 
design. The height of the bell and the diameter at its base 
vary from if to 2 times the chimney diameter. The bell sec- 
tion is usually made of a number of pieces, butt jointed with 
single outside straps as shown in the drawing. The bell sheets 
are usually -jjpin. thicker than the ring sheets immediately above 
them. 



262 



GRAPHICS AND STRUCTURAL DESIGN 



Lining. — The method of lining varies. Where the tempera- 
tures do not exceed 650 F. (and they 
ordinarily do not) common red brick 
are satisfactory; otherwise a No. 2 
fire brick is required. At the top the 
lining should have a thickness 4I ins., 
which can be increased by 4J ins. 
every 30 to 40 ft. In some cases the 
chimney is only partially lined, say 
one-half the way up, and in other 
cases the thinnest lining is made 2\ 
ins.; the former, however, is the bet- 
ter practice. 

Where the breeching from the boil- 
ers enters the chimney above the base 
care should be taken not to weaken 
the chimney by cutting away too 
much, and the chimney should be 
sufficiently reinforced at this point. 
The practice of having the gases enter 
the chimney through flues below the 
steel base makes the chimney stronger 
and of better appearance. 

Foundations 

The shell is maintained upright 
against the moment of the wind pres- 
sure by the weight of the foundation 
and chimney with lining assumed as 
acting about the lower edge of the 
foundation. 

The foundation is concrete and the 
chimney is secured to it by heavy 
foundation bolts. The concrete can 
be assumed as weighing from 125 to 150 lbs. per cu. ft. When 




CHIMNEYS 263 

there is no wind acting upon the chimney the combined weights 
of the foundation, W f , the steel chimney, W c , and the lining, 
W t , are distributed uniformly over the soil through the base of 
the foundation, as is shown in Fig. 277 (b). Now as a wind 
pressure F begins to act the uniform pressure previously on the 
soil will be altered, the pressure being reduced on the side upon 
which the wind acts and increased on the opposite side. In 
Fig. 277 (b) 5 and W f + W c + W h which are always equal, He 
in the same line. In Fig. 277 (c) the force acting upon the 
base from the soil, S, has moved over, acting through the center 
of gravity of the trapezoid which here represents the distribu- 
tion of the soil pressure. 

When the wind pressure becomes great enough the distribu- 
tion of pressure on the soil will become that shown in Fig. 277 (d), 
being zero at the right, while the original pressure at the left is 
doubled. The force S will pass through the center of gravity 
of the triangle, or J B from the edge of the foundation. It is 
commonly assumed in masonry construction that the resultant 
pressure upon any section must pass within the kern of that 
section. Were the resultant S to pass to the left it would pass 
out of the kern. See page 247. In Fig. 277 (a) the moment 
due to the wind force must equal the moment of the forces W f) 
W c and Wi, with the arm a; hence 

F@f + H^j = {W f + Wc + Wi) a; * 

the limit being a = 0.118 B we have 

F (^ + H f ) = (W f + W C + W t ) X 0.118 B. 

The volume of a frustum of a cone or pyramid is 



h 



V = (A t + A b + VA t XA b )-, 

3 
where A t = area of the top of the frustum. 

Ab = area of the bottom of the frustum. 

h = altitude of the frustum. 

* Note. — Some designers neglect the weight of the lining, as it may be omitted 
or removed during the life of the chimney. 



264 GRAPHICS AND STRUCTURAL DESIGN 

Maximum Pressure on the Soil. — When the resultant pressure 
passes through the kern the maximum pressure will be less than 
twice the mean pressure. The maximum pressure per square 
foot then is 

A b 

P = pounds per square foot. 

A b = area of the base of foundation, square feet. 

Where the foundation is not reinforced the angle of the sides with 
the vertical should not exceed 30 degrees. 

The maximum pressure should not exceed that permitted on 
the soil (see page 236). 

In the chimney under discussion assume the base 18 ft. square 
and that the concrete weighs 140 lbs. per cu. ft. Estimating the 
weight of the steel of the chimney, the diameter being 5 ft. 6 ins., 
the circumference is 17.3 ft. 

Sheets. 

A 

1 
4 

A 



Allowing 10 per 



Dimensions, 


Weights, 


feet. 


lbs. 


17.3 x 30 x 10.2 x ! 


3,980 


I 7-3 X 30 X 10.2 X 1 


5,300 


!7-3 X 30 X 10.2 X f 


6,620 


17.3 X 10 X 10.2 X f 


2,650 




18,5501 


for bolts, rivets, laps, etc., 


1,850 



Total 20,400 lbs. 



The weight of the lining assuming the chimney to be lined to 
a height of 62 ft., 



Section. 


Lining, 


Area of lining, 


Volume, 




ins. 


sq. ft. 


cu. ft. 


Base to 40 ft. 


9 


II. 2 


448 


40 to 62 ft. 


4* 


6.05 


135 



Total 583 

Weight of lining, 580 X 125 = 72,500 lbs. 

The height of the foundation will be assumed at 10 ft. 



CHIMNEYS 265 

If the resultant of wind pressure and total weight is to fall 

within the kern its distance from the center line of the chimney 

must not exceed 0.118 X B = 0.118 X 18 X 12 = 25.5 ins.; 

this is the distance along the diagonal. At right angles to the 

h 12 

side of the square it would be - = 18 X -7- = 36 ins. 

6 6 

Trying the following foundation, top 10 ft. square, bottom 

18 ft. square and height 10 ft., the volume of this foundation 

is 

V = - (A t + A b + VA t xA b = — (10 2 + 18 2 + Vio 2 X 18 2 ) 
3 3 

= 2013 cu. ft. 

Weight = 2013 X 140 = 281,800 lbs. 
The total weight on the soil is 281,800 + 20,400 + 72,500 

= 374,700 lbs. The uniform pressure per square foot is ^' ' — 

3 2 4 
= 1150 lbs. The resultant of wind pressure and total weight 
cuts the bottom of the foundation a distance q from the center 
line of the chimney. The total wind force is F = 5.5 X 100 X 30 
= 16,500 lbs. This force acts at a distance of 50 ft. from the 
base of the chimney or 60 ft. from the bottom of the foundation. 

„ 60 X 12 X 16,500 

Hence a = ■ — — = 31.7 ms. 

374,7oo 

The resultant evidently falls outside the kern on the diagonal 
but passes within it at right angles to the side. 

The maximum pressure on the corner can be found by the 
assistance of the curves, Fig. 269. 

P h 18 X 12 47 

From the curves for a square base when /? = 0.147, a = °- 22 > 
and since 



266 



GRAPHICS AND STRUCTURAL DESIGN 



the foundation will be satisfactory if this load of 2625 lbs. per 
sq. ft. is permissible upon the soil under the chimney. 

Foundation Bolts 

The chimney is assumed as tending to overturn about the 
axis a-a tangent to the bolt circle and differing only slightly 
from the lower edge of the bell. 
The maximum fiber stress will 
be f m and will occur in bolts 
No. 3 and No. 4 (see Fig. 278), 
these being farthest from a-a. 
The fiber stress in any other 
bolt will be f m multiplied by 
the ratio of its distance from 
a-a to the distance of the bolts 
farthest from a-a; thus the 
fiber stress in the bolts 2 and 

5 is f m X - . The resisting mo- 

ment due to any bolt is the product of its area, the fiber stress 
acting in it and its distance from the axis a-a. The resisting 
moment of the entire group of bolts will be the sum of the 
moments of the individual bolts. Taking the case of the six 
bolts shown, and letting a be the area of one bolt, at the root 
of the thread, and R the radius of the lower bell circle, we have, 




Fig. 278. 



Fiber stress. 



Moment. 



No. bolts. p 

iand6 R (1 - cos 30 ) = 0.134^ fmX °'l^t 2-a-fm-R — 

1. 000 K I. 



O.134 2 



866 
1 
1.866 



2 and 5 R (1 — cos oo°) = R f m X Q ,, p 2-a-f m *R 

1. 000 K 

3 and 4 R (1 - cos 150°) = 1.866 R f m Xi 2 • a -fm • R • 1.866 

In a similar manner take moments around one of the bolts 
and use the lower of the two values. 

This gives a total moment of M = 4.5 X a -f m • R. 

In this way it may be shown that in the expression M = 
C ' a • f m • R, C equals three-quarters of the number of bolts. 



CHIMNEYS '267 

The moment on the chimney about its base due to the wind 
is 8,250,000 in. lbs. The moment due to the weight of the 
chimney which tends to hold it upright is M\ = 20,400 X 45 
= 918,000 in. lbs. The net moment to be resisted by the bolts 
is 8,250,000 — 918,000 = 7,332,000 in. lbs. 

Trying six bolts, M = 4.5 a -f m R, or a = 7>33 2 ,ooo = 

4.5 X 10,000 X 47 
3.46 sq. ins. This corresponds to 2j-in. diameter bolts. This 
method being approximate, low fiber stresses should be used in 
bolts and brackets. Fig. 276 shows an effective bolt-bracket. 
The bolts should be kept close to the edge of the bell, and the 
bell sheet reinforced with straps. The lower edge of the bell 
should be reinforced with a circular band. This lower edge then 
rests upon a cast-iron base plate, Fig. 273, that distributes the 
pressure over the top of the foundation. The base plate may be 
made in one piece for small chimneys, as shown, but is usually 
made of a number of sections for convenience in casting and han- 
dling. These sections when placed in position are secured with 
a few f-in. diameter bolts. After erection the bell of the chimney 
and the foundation bolts will hold these sections in place. 

The foundation bolts should extend almost to the bottom of 
the foundation and should be fitted with a lock nut and washer 
or key and washer. 

The chimney should have a light ladder running to a platform 
at the chimney top. Dropping the platform a little below the 
top protects the handrail from the chimney gases. 

A somewhat more conservative treatment might consider the 
pressure acting upon one-quarter of the area of the bell rim 
rather than being concentrated upon its leeward edge. The 
center of pressure might then be considered as acting on the 
leeward side, nine-tenths of R from the bell center. R is the 
radius of the bolt circle which has been considered as being about 
the center of the bell rim angle. Under these conditions C 
equals two-thirds of the number of bolts. 



268 GRAPHICS AND STRUCTURAL DESIGN 

Reinforced-concrete Chimneys 

In a reinforced-concrete chimney the weight of the concrete 
above the section under consideration will produce a compression 
on that section similar to that produced in a masonry chimney. 
Take any unit area having a ratio of reinforcement of p ; under a 
given load the shortening of the two materials will be equal and 
the loads shared proportionally to their moduli of elasticity. 

Let/ C = compressive unit stress in concrete, pounds per square 

inch. 
p 
— - = n; then the fiber stress on the steel is f a = n >f e and 

if L is the load on i sq. in. of section/, = -. -. The 

i + p (n - i) 

load can be assumed as the weight of the concrete for the ordi- 
nary percentages of steel reinforcement. 

With wind acting on one side of a chimney the pressure upon 
the leeward side is increased while that on the windward side 
is decreased, finally becoming zero while that upon the leeward 
side becomes double the original direct unit pressure. The 
concrete not being assumed as resisting tension may however 
be considered as acting like a beam until the compression at the 
windward side becomes zero. Under these circumstances the 
resistance of the section may be used as in the case of a beam 
resisting both tension and compression. In calculating the 
moment of inertia of the section proper allowance must be made 
for the modulus of elasticity of the steel differing from that of 
the concrete. 

I = I c + /. = 4 C° 2 + d ^ + °'394 n • D 1 H S (A -3 Q; 
10 

here, / = total inertia of concrete and steel. 

I c = inertia of concrete ring. 

/, = inertia of steel. 

A = area of concrete ring, square inches. 
D = outside diameter of section, inches. 



CHIMNEYS 



269 



d = inside diameter of section, inches. 
D\ = diameter of steel cylinder having same area as total 

reinforcement at the section. 
t 8 = thickness of equivalent cylindrical steel reinforcement. 

Generally t 8 will be so small that 3 • t a may be neglected, making 



/ = ^(Z) 2 + d 2 ) +0.394 nDt% 
10 

Where the ratio of reinforcement is known t a = - (D — d). 

2 

When the point has been reached where the extreme fiber 
stress on the concrete on the windward side is zero any further 
increase of the wind pressure F will be resisted by increased 
compression in the concrete on the leeward side and by the re- 
inforcing steel taking tension on the windward side. This causes 
the axis of zero stress to move across the section from the extreme 
fibers on the windward side towards the leeward side. Consider- 
ing now only the additional moment applied after the extreme 
fiber stress on the windward side becomes zero there will be 
created additional compression on the leeward side shown 
shaded, while on the windward side the fiber stress will be tensile 
and must be carried by the steel. Con- 
sidering now the fiber stress as indicated 
in Fig. 279, it will vary uniformly on 
each side of the neutral axis a-a, the 
stress in the steel being n times the stress 
in the concrete at the same distance from 
the axis a-a. 

The resultant of the flange stresses in 
compression, shown in the shaded por- 
tion in Fig. 279, must equal the resultant 
of the tensile stresses in the steel in the 
remaining part of the section, the portion 
resisting moment induced in any section 




nSW 



Fig. 279. 

not shaded. The 

equals either flange force multiplied by the distance between 



270 GRAPHICS AND STRUCTURAL DESIGN 

the lines of action of the resultant flange forces on each side of 
the axis a-a. This distance, where R = R 8 , has been found to 
be constant and to equal 1.56 R s , where R s = radius of the steel 
circle. 

The values of the flange forces and the moments have been 
calculated based upon the assumption that the stresses vary pro- 
portionally to the distances of the fibers from the neutral axis 
a-a, and that the fiber stress in the steel is n times that in the 
concrete for similar positions. The thickness of the wall Ri — r 
(Fig. 281) has been considered small compared with R; that is, 
the flanges have been considered as lines. The flange forces are 

Tensile steel, F t = 2 t s f t R s X C\. 
Compression steel, F c = 2 t s fcR s X C 3 . 
Compression concrete, F c0 = 2J co t C oR X C 3 . 

The moments due to these forces are 
Tensile steel, M t = 2f t t 8 R 8 2 X C 2 . 
Compression steel, M c = 2f c t s R s 2 X C 4 . 
Compression concrete, I M = 2fcot C0 R 2 X C 4 . 

Here t s = thickness of equivalent cylindrical steel reinforcement, 

inches. 
ft = extreme fiber tension in steel, pounds per square inch. 
R s = radius of steel, inches. 
f c = extreme fiber compression in steel, pounds per square 

inch. 
t C o = thickness of concrete, inches. 
fco = maximum compression in concrete at mean radius R, 

pounds per square inch. 
R = mean radius of concrete, inches. 
f a = fiber stress in the concrete at the radius R s . 

The values of these coefficients G and C 3 have been calculated 
for the several values of k and are given in the curves, Fig. 280. 
The following relations are important. F t = F c + F c0 . The 
total resisting moment of a section M = M t = M c + M c0 . 
The distance between the lines of action of the resultants of the 



CHIMNEYS 



271 



flange forces has been previously stated as 1.567?,; hence the 
moment may be expressed as 

M = F t X 1.56 R = (F c + F„) 1.56.12. 
These curves are based upon n = 15. 





1.51 

l.A( 

1.3C 

La 

l.K 
l.fll 

.8ft 

.70 

..60 

,50 

.±0 

.30 

.20 

.10 
• 1 


1 




RE-IN.FORCED CONCRETE 
CHIMNEYS 












110 

130 
120 
110 
100 
090 
OSO 
070 
000 
050 
010 
030 
020 

m 




\ 




F= 2 tsft^sti 






















\ 




M=Fxl.58R s 






















\ 






























\ 






























\ 




Ci 


























\ 
































if" 






























\ 






























\ 




























^. 


Jl 































V 


** 
























1 




\ 






^r Cs 




















03 
O 

O 

<D 

1 




\ 






























vz 




























V 




























3: 






























_s 
















































































































































































































































Atsc 


ale on 


lefttc 


* 
























iU 


multi 


jUedl 


ly 100. 








^r p 










































3 

L1_UL1_ 


,M! 2 


',,!, 


m? 3 





A 


) 


•S 


, f M 


rn', 6 


L, 


, * . ; 7 








Scale of k. 
FlG. 280. 

The curves also give the relations between the several fiber 
stresses and the critical percentages of reinforcements for the 

various values of k. k = — . Here R is the radius of the flange 

R 

being considered, that is, R 8 for the steel and the mean radius 
for the concrete. The following problem will illustrate the cal- 
culations of a chimney as outlined. 

* Note. — The values of -j and p have been estimated upon the basis 

fa 

that R = R s . This materially simplifies the problem. 



272 



GRAPHICS AND STRUCTURAL DESIGN 



Problem. — A chimney is 150 ft. high. Its outside diameter 
is 10 ft. and the walls at its base are 9 ins. thick. Assume a 
wind pressure of 50 lbs. per sq. ft., giving an equivalent pressure 
of 30 lbs. per sq. ft. of projected area of the round chimney. 
The pressure on the concrete is not to exceed 600 lbs. per sq. in., 
while the unit stress on the steel must not be over 15,000 lbs. 
per sq. in. The chimney walls are 6 ins. thick for the upper 
100 ft. and 9 ins. thick for the lower 50 ft. In Fig. 282 assume a 
reinforcement of 1.75 per cent and that a bar of concrete 1 sq. 





Fig. 281. 



Fig. 282. 



in. in section and 12 ins. high weighs 1 lb. The direct compres- 
sion in the concrete at the base of the chimney is 

r (100 X i) -h 50 117 1U 

f c = - , 9J \ = — '- ~ 100 lbs. per sq. in. 

1 + p (n — 1) 1.24 

The inertia of the section with 1.75 per cent steel is 

„ A 



16 



(£> 2 + d 2 ) + 0.394 nD x \. 



j = 3_i4°. ( I20 2 + IQ2 2) + 0#394 x 15 X 114 3 X 0.158 
16 

= 6,252,000 in. 4 . 

t,- fl IOO X 6, 2^2, OOO . „ 

M = — = -^- 2 = 10,420,000 m. lbs. 

e 60 

The total wind force is W = 10 X 150 X 30 = 45,000 lbs. The 

12 
moment due to this wind pressure is M = 45>o°° X 150 X — 



CHIMNEYS 273 

= 40,500,000 in. lbs. The remaining moment to be resisted by 
the reinforced section after the stress in the extreme fibers on 
the windward side has become zero is 40,500,000 — 10,420,000 
= 30,080,000 in. lbs. The compression on the leeward side at 
this time is 2 X 100 = 200 lbs. per sq. in. The permissible in- 
crease in the compression on the leeward side is 600 — 200 = 
400 lbs. per sq. in. 
The additional flange force is 

„ 30,080,000 30,080,000 , „ 

F = ? ' ' . = -^- 2 - — l = 346,000 lbs. 

(1.56X2O 1.56X57 

Ft = 2 X t 8 Xft X R. X G = 2 X 0.158 X 15,000 X 57 X 1.25 

= 338,000 lbs. 

This is sufficiently near the required 346,000 lbs. From the 
curves, Fig. 280, the ratio of stress in the steel to that in the con- 
crete is 41 to 1 ; hence 

, 15,000 „ 

fc = -*" = 365 lbs. per sq. m. 

4 1 

The approximate total stress in the concrete = 200 + 365 
= 565 lbs. It should be noted that the 365 lbs. is the flexural 
stress at a distance R s from the center of the chimney and that 
this can be easily corrected. 

The value of k from the curves corresponded to 0.46, making 

A = k X R s = 0.46 X 57 = 26.2 ins. 

The distance from the neutral axis to the steel 57 — 26.2 = 30.8 
ins. The distance to the outside of the concrete is 60 — 26.2 
= 33.8 ins. By similar triangles, 

% 5 = 2i f, or / max = 4 oolbs. 

The extreme fiber stress then becomes 200 + 400 = 600 lbs. per 
sq. in. 



274 GRAPHICS AND STRUCTURAL DESIGN 

Design of a Reinforced-concrete Chimney 

In Fig. 283, assuming the chimney wall 5 ins. thick at the top, 

the area of this section is 12 10 sq. ins. The radius of the kern 

is 

r 



D( .d 2 \ 82/ , 7 2 2 \ . 

8( 1 + ^j = y( 1 + 8?j = l8 -3 ins - 



This means that before any reinforcement would be required the 
resultant of wind force and weight could pass 18.3 ins. beyond 
the axis of the chimney. It follows that the weight above a 
section I feet from the top, divided by the wind force acting on 

the portion above that section equals -ins., divided by 18.3, or 

2 

I 

1210 X / 2 , 1210 X 18.3 X 2 ,, 

■ = -—— . or I = — — ^ = 18. 1 ft. 

30 X 6.8 X / 18.3 30 X 6.8 X 12 

Hence the upper 18 ft. would require no reinforcement. It is 
usual to place a small reinforcement in this portion. Assuming 
a reinforcement used here of \ per cent (0.005) the distance that 
this will serve down the chimney can be found in a similar way. 
From the curves the ratio of the fiber stresses for p = 0.005 * s 
85 to 1 ; hence taking the fiber stress in the steel at 15,000 lbs. per 

sq. in. would make the value of f e = -^ = 177 lbs. per sq. 

in.; as this is very low the tensile steel will determine the strength 
of the section. The value of C± is 1.38. The flange force in 
the tensile steel then is 

F t = 2 tj t X R a X Ci = 2 X 0.025 X 15,000 X 38.5 X 1.38 

= 39,800 lbs. 
Mi = F t X 1.56 XR 8 = 39 7 8oo X 1.56 X 38.5 = 2,395,000 in. lbs. 

The inertia of the section is 

10 

I = ^r (82 2 + 72 2 ) + 0.394 X 15 X 77 3 X 0.025 = 968,406. 
10 










% VV 



Fig.. 288 



REINFORCED 
CONCRETE CHIMNEY. 



(275) 



276 GRAPHICS AND STRUCTURAL DESIGN 

Now M = — , and when the fiber stress on the windward side 
e 

becomes zero that on the leeward side has become double the 

stress due to the weight above the section. 

If the concrete weighs 144 lbs. per cu. ft., and the distance 

from the top to the section is I feet, then f = I, and we have 

,, II I X 968,406 , 7 

M 2 = — = — — — = 23,600/. 

e 41 

The bending due to the wind acting on the portion above the 
section equals the sum of M x and M 2 . From the wind pressure 
and the chimney dimensions the bending equals 



M- 



30 X 6.8 X I 2 X 12 72 

= - = 1220 P. 



3 — 



Equating these values of M we have 

1220 X I 2 = 23,600 X / + 2,395,000, 
from which 

I 2 - (19.3 X I) + (^J = 2058, or I = 55.0 ft. 

This reinforcement will have an area of 12 10 X 0.005 = 6.05 sq. 
ins., and will require 14 — f-in. round bars. Allowing a bond 
stress of 80 lbs. per sq. in. the length of the bars at laps is 

h = 7 — ? 

where h = length of the lap in inches. 

f 8 = fiber stress in steel, pounds per square inch. 
f b = bond stress, pounds per square inch. 
d = diameter of round bar or side of square bar, inches. 

In the problem 

, 15,000 X d , , 

k = ^^ = 46 - gd ' 

d being f in. the lap should be 46.9 X 0.75 = 35.2 ins. 



CHIMNEYS 277 

The reinforcement will now be determined for the section at the 
base. The moment due to wind at this section of the chimney 
is 

M = [(85 X 6.8 X 30) (^ + 40)] + [(40 X 8.5 X 30 X 20)] 

= 1,634,550 ft. lbs. 
M = 19,614,600 in. lbs. 

D = 102 ins. d = 88 ins. The area of the section is 2085 sq. 
ins. The weight of the upper 85 ft. is 12 10 X 85 =.103,000 lbs. 

The load per square inch on the lower section is/ c = — **- — -f 40 

2085 

= 50 -f 40 = 90 lbs. per sq. in. Trying a reinforcement of 
0.0175, from the curves d = 1.25, y = 41, and k = 0.46. 

Ja 

Revising f c for the steel percentage 

,. 90 90 

Jc " 1 + p (n - 1) ~ 1 + (0.0175 X 14) ~ 72 ' 

The compression on the leeward side when the tension on the 
windward side is zero is 2 X 72 = 144 lbs. per sq. in. The 
tension in the steel corresponding to this is (500 — 144) 41 = 
14,596 lbs. per sq. in. 

F = 2 tJtRsCi = 2 X 0.1225 X 14,600 X 48 X 1.25 

= 214,600 lbs. 
M = F X 1.56 X R s = 214,600 X 1.56 X48 = 16,069,250 m. lbs. 
The inertia of the section is 

/ = 4 (D 2 + d 2 ) + 0.394 XnX D x % 
10 

= ^ (102 2 + 88 2 ) + 0.394 X 15 X 96 s X 0.1225, 
10 

I — 2,365,000 + 640,000 = 3,005,000, 

1 , /7 72 X 3,005,0 00 

M = — = L — — — = 4,240,000. 

« 5 1 



278 GRAPHICS AND STRUCTURAL DESIGN 

The remaining moment to be resisted by the chimney is 

19,614,600 — 4,240,000 = 15,374,600 in. lbs. 
F = M -^ (1.56 X R 8 ) = i5,374,6oo ^ (1.56 X 48) = 205,000 lbs. 
ft = F + (2Xt 8 XR 8 XC 1 ) = 205,000 -f- (2 X 0.1225X48X1.25) 

= 13,900 in. lbs. 
/. = liSoo _ 34Q lbs _ 
4 1 

The extreme fiber stress will be somewhat in excess of this. 
k = 0.46, A = k X R s = 0.46 X 48 = 22 ins. 

26 ~\- 7, 

By similar triangles the extreme fiber stress is — — X 340 

26 

= 378 lbs. per sq. in. The extreme fiber stress then becomes 

(2 X 72) + 378 = 522 lbs. per sq. in. The reinforcing bars 

will require an area of 2085 X 0.0175 =36.5 sq. ins. Using 

^6 ^ 
J-in. bars will demand ^-^ = 8^ — f-in. bars. 

0.44 

Similar calculations will indicate the following bars in the 

several sections; beginning at the bottom, we have 

Section. Section. 

First, 22 ft., 83 f-in. round bars. Fourth, 22 ft., 28 f-in. round bars. 

Second, 25 ft., 60 f-in. round bars. Fifth, 34 ft., 14 f-in. round bars. 

Third, 22 ft., 44 f-in. round bars. 

Horizontal reinforcing rings must be used to resist the web stresses 
and also to prevent cracking due to expansion from the heat. 
Concrete being a poor conductor of heat the inside becomes much 
hotter than the outside. This causes considerable circum- 
ferential stress in the chimney which must be resisted by hori- 
zontal rings, preferably placed near the outer surface of the 
chimney. 

The amount of this circumferential reinforcement can be only 
roughly approximated, as the difference in temperature between 
the inner and outer faces of the wall must be guessed. In 
several chimneys this reinforcement was from J to f of i per cent. 



CHIMNEYS 



279 



The greater the steel ratio the higher is the compression in the 
concrete. 

These chimneys are generally lined at least one-third their 
height. The inner tubes, being protected from the wind, are 
merely called upon to resist temperature stresses. These are 
not usually calculated. The linings may be made of fire brick 
or reinforced concrete and will have horizontal ring reinforce- 
ment somewhat lighter than in the outer wall, say from J to J 
of 1 per cent, and vertical reinforcement to resist temperature 
stresses of from f to \ of 1 per cent. 

The horizontal reinforcing rings are commonly |-in. or f-in. 
rounds, and are spaced from 12 ins. to 18 ins. vertically. The 
spacing should be closer at the point on the chimney where the 
lining stops, Fig. 286, and additional vertical reinforcement should 
also be placed at this part. 

An approximation of the amount of steel for web reinforcement 
maybe made following the formulae given on page 218. As- 
suming no tension carried by the concrete we find 

wXH 
^~i8.7 Xtco Xft 

Here w is the effective horizontal wind pressure, in pounds per 

square foot. 
H is the height of the chimney above the section being 

examined, in feet. 
t Co is the thickness of the chimney wall at the section, in 

inches. 
ft is the unit tensile working stress on the steel, in pounds 

per square inch. 
p is the ratio of steel reinforcement. 

To this calculated percentage there should be added a small 
amount of steel to provide against vertical temperature cracks. 
Taylor and Thompson place this amount at \ of one per cent, 
and recommend the placing of these horizontal rings at intervals 
of from 6 to 10 inches, rather than at 12 inches or over. 



Area, 


Volume, 


Weight, 


sq. ft. 


cu. ft. 


lbs. 


W 1 -?- x 85 


= 715 X 144 = 


103,000 


- 2 tW X 40 


= 580 X 144 = 


83,SOO 


Iff X 40 


= 265 X 144 = 


38,200 


56 X 3 


= 168 X 144 = 


24,190 




= 


167,000 




Total 


415,890 



280 GRAPHICS AND STRUCTURAL DESIGN 

Chimney Base, Fig. 288. — Estimate of total weight of 
chimney : 

Section. 

Upper 

Lower 

Lining 

Base 

Foundation 



The total wind moment previously found for the base of the 
chimney is 1,634,550 ft. lbs. The total force is (85 X 6.8 X 30) 
+ (40 X 8.5 X 30) = 27,540 lbs. The resultant wind pressure 

then must act ' = kq.k ft. above the ground line. The 

27,540 

resultant of weight and wind will act a distance q from the axis 
of the chimney and 

(59-5 + 7) P 66-5 X 27,540 ,^ . 

'? = ur = — q"^ = 44ft. = 52.8 ms. 

The radius of the kern for a square is r = 0.118 X h = 0.118 
X 20 X 12 = 28.4 ins. From page 255, 

0.22, and a = 0.173, 





h 20 


since 




W = 


2af e h 2 , or f e = 



W ' 89 ° , = 3000 lbs. 
2 ah 1 ' 2 X 0.173 X 20 1 

<j> = 0.27. The resultant pressure passes 4.4 ft. from the cen- 
ter, so that the pressure extends a distance (5.6 X 3) = 16.8 ft. 
over the base. The area of the pressure on the base is 16.8 X 20 
= 336 sq. ft. The total pressure equals the entire weight of 

the chimney, hence 336 X - = 415,890 and 

2 

P = (415)890 X2)t 336 = 2470. 

The reinforcement can be only approximated, the simplest way 
being to consider the part cdef, Fig. 288, as a can tile vered beam. 



CHIMNEYS 281 

The distance from the center of gravity of the portion cdef to 
the line cf is 

Area. Statical moment. 

+ 20 X V- = 100 X - l f = 333-33 

- 10 X f = _£5 X (5 + f ) = 166.75 
75 166.58 

_. 166.58 

X = — = 2.22 ft. 

75 

The load on cdef approximates 75 X 2400 = 180,000 lbs. The 
approximate bending moment is M = 180,000 X (5 — 2.22) 
= 500,400 ft. lbs. The bending moment per foot along the 
line de is (500,400 X 12) -r- 10 = 600,480 in. lbs. Assuming 
n = 15; d = 42 ins. and p = 0.0025; j = 0.922 and kj = 0.218. 
Using the formula for reinforced concrete design, we have 

, _ M 8 600,480 

A Xjd (42 X 12 X 0.0025) X 0.922 X 42 
= 12,300 lbs. per sq. in. 
and 

, 2 X M c 2 X 600,480 , „ 

fc= t~. TTo = n ; = 26 ° lbs. per sq. in. 

. kj X bd 2 0.218 X 12 X 42 2 v h 

The depth of the beam being controlled by the necessity of the 
foundation reaching proper soil and extending below the frost 
line this fiber stress in the concrete is satisfactory. The spacing 
of if-in. round bars will be 

area of bar X 12 0.004 X 12 , . 

spacmsr = : = ^^ ■ = o4 ms. 

& A 42 X 12 X 0.0025 V2 

The subject of stirrups for the web stresses may be considered 
as suggested on page 218. 



CHAPTER XVII 
RETAINING WALLS 

Pressures on Retaining Walls 

According to Coulomb's theory some wedge BAC, in Fig. 289, 
will produce a maximum pressure E against the wall. This 
force will make an angle 8 with the face of the wall corresponding 




Fig. 289. 



to the angle of friction between the face of the wall and the fill. 
The natural slope of the fill is the angle <j>\ the angle that the 
face of the wedge, producing the maximum pressure against the 
wall, makes with the horizontal is x. 
h = height of wall in feet. 

w = the weight of the fill in pounds per cubic foot. 
According to this theory the general value of the maximum 
pressure in pounds per foot of length of wall is 

sin 2 (0 - </>) 



w • h' 



. 2/1 • (a l*\/ 1 4 / sm (0 + h) sin (4> - a) Y 

sm 2 • sin (0 + 5) ( 1 + V . >„ , »( . > r 

v J \ V sin (0 + 5) sin (0-a)/ 



282 



RETAINING WALLS 



283 



This is simplified for the more commonly assumed conditions 
as follows : 

For = 90 degrees, and <5 = a, 



E = -wh 2 

2 



COS 2 



COS 



a(l+V/" 



sin (<f> -+- a) sin (<f> — a) 



InRankine's formula = 90 , a = o, and 5 = o; substituting 
these values in the general equation gives : 
„ I 7 2 cos 2 <f> 

2 (1 + sm0) 2 
£ = I w • h 2 tan 2 (45 - %<f>). 
„ 1 y 2 1 - sin 
2 i + sin cf> 

This theory, although usually developed for retaining walls, 
applies also to the sides of bins. 




Fig. 290. 

Graphical Solution. — The graphical method (Fig. 290) affords 
the neatest way of estimating the pressure E. 

E = iw-p.y. 



284 GRAPHICS AND STRUCTURAL DESIGN 

p and y = lengths of lines in the diagram measured to the same 
scale by which the wall or side is drawn. The other symbols 
are the same as in the preceding case. 

Explanation of Diagram. — A B is the side of the wall or bin 
in contact with the fill. A I is a horizontal line through the base 
of the wall or side AB. AC is the line of natural slope of the 
fill making an angle </> with the horizontal. BC is the slope of 
the top of the fill making an angle a with the horizontal. Pro- 
duce AC and BC until they intersect in C. Upon iCas a diam- 
eter draw the semicircle ARC. Draw BG making an angle 
+ 8) with the wall AB. 

At the point of intersection G of BG and AC draw GH perpen- 
dicular to AC. With AH as a radius draw the arc HF and 
through F draw DF parallel to BG. From D drop p perpendicu- 
lar to AC. 

The proofs and discussions of the preceding equations and 
diagrams may be found in books on the mechanics of retaining 
walls and earth pressures. 

Figure 291 is the diagram when the angle of fill a equals the 
angle of natural slope <£. In this case since BK and AF are 
parallel p and y will be the same lengths wherever the point D 
is taken in the fine BK. 

Figure 292 shows the construction when the angle of fill falls 
below the horizontal and also below the line BC 1 making an angle 
4> + 8 with the back of the wall. In all the preceding cases 
E = \ w • p • y. 

If the fill, Fig. 293, carries a uniform load L lbs. per sq. ft. of 
horizontal projection, the height of the fill may be considered 
as increased sufficiently to bring such loading on the soil. The 
diagram is given in Fig. 294. 

w' = w H r~ and E = -\w -\ — )£*y. 

h 2 V h I 

Distribution of Pressure on the Wall or Side. — In the first 
cases, where the fill is not loaded, the pressure on the wall will 



RETAINING WALLS 



285 




Fig. 293. 



Fig. 294. 




Fig. 297. 



286 



GRAPHICS AND STRUCTURAL DESIGN 



vary as a triangle and the center of pressure will correspond to 
the center of gravity of the triangle, as shown in Fig. 295. 

Where the fill is loaded the usual assumptions regarding, the 
distribution of the pressures behind the wall are shown in Fig. 296 
and the magnitudes of these pressures are given by the follow- 
ing formulae. The resultant pressure E t acts through the center 
of gravity of the trapezoid of pressure abed. 

*-a( w+ ir)-'-* and M^t^ 1 - 



From which E" 



2E t 

h 



-E\ E' = 



L-p >y 

hxk 



Center of Gravity of a Trapezoid. — Fig. 297 being a trape- 
zoid its center of gravity can be readily found as shown. 



WEIGHTS AND ANGLES OF REPOSE OF MATERIALS 



Material. 



Weight per cu. ft. 



Angle of repose, <f>. 
Degrees. 



Clay, dry , 

Clay, damp 

Clay, wet , 

Gravel, wet 

Ashes 

Coke 

Earth, dry 

Earth, moist 

Earth, wet 

Broken stone, wet 

Coal, broken 

Sand, dry 

Sand, moist. ...... 

Sand, wet 

Water 



90-110 
100-120 
120-135 
100-120 

40-45 

30 

80-90 

90-100 
105-120 
100 

56 

90-110 
100-110 

115-125 

62.5 



30-40 
40-45 
15-25 
25-40 
25-40 
30-45 
30-40 

35-45 
17-30 

35-40 
45-50 
30-35 
30-45 
15-30 

o 



The coefficient of friction between the usual masonry materials 
upon themselves or upon soil will range from 0.50 to 0.75. 



Note. — To prevent the fill becoming saturated with water and greatly 
increasing the pressure on the wall the fill should be carefully drained. 
This may be done by suitably locating a drain of broken stone or gravel 
behind the wall and connecting at intervals by weepers draining outside 
the wall. 



RETAINING WALLS 



287 



Retaining Wall 

A retaining wall may fail by being rotated about the toe A or by 
sliding upon the base AB. There is usually little likelihood of 
failure by sliding as the coefficient of friction between the wall and 
the soil is high. The force E due to the earth pressure behind the 
wall creates an uneven pressure under AB as shown in Fig. 298. 

A tilting of the wall may result from the side at A settling 
faster than that at B. To improve the condition it is not 
necessary to enlarge the entire wall but the base may be spread 
as shown in Fig. 299, so that the resultant R passes through the 
center of the base AB. 




Fig. 298 



Fig. 299. 



At best, the theories relating to retaining walls are unsatis- 
factory. This is due largely to the fact that the properties of the 
materials vary so widely that it is difficult to decide upon the con- 
ditions applicable to a particular case. The practical conditions 
differ greatly from the approximations necessary in the theory. 
The weight of the fill and its angle of repose will vary greatly 
as the fill is wet or dry, clean or dirty, loose or rammed, etc. 

Other factors, such as shock, unexpected loading and frost, 
cannot be taken into account in the formulae. According to 
Trautwine a practically vertical retaining wall sustaining a fill 
of sand, gravel or earth, without surcharge, with the fill loose, 
not rammed, the wall being of good common rubble or brick, 
should have a thickness at the top of the footing of four-tenths 



288 GRAPHICS AND STRUCTURAL DESIGN 

the height of the wall. The Engineering News in its issue of 
Sept. 26, 191 2, commenting upon the failure of a retaining wall, 
recommends the calculation of a retaining wall upon the assump- 
tion that a wall with no surcharge carries a load at least equiva- 
lent to a mass of water behind the wall of two-thirds its depth. 
The following problem will illustrate the method of designing a 
retaining wall. 

Problem. — Design a section for a retaining wall, height 24 ft., 
weight of earth no lbs. per cu. ft., natural slope 30 degrees, 
maximum earth pressure not to exceed 5000 lbs. per sq. ft. 
Assume that the concrete of the wall weighs 140 lbs. per cu. ft. 

The center of gravity of the wall must be located with reference 
to the back of the wall. Dividing the wall section (see Fig. 
300) into rectangles and triangles, multiplying their several 
areas by the distances of their centers of gravity from the back 
of the wall and then dividing the sum of these moments by the 
total area of the wall section gives the center of gravity of the 
entire wall section as 46.6 ins. from the back of the wall. The 
area of the wall section is 143 sq. ft. 

The weight of 1 ft. of length of the wall is 143 X 140 = 20,020 
lbs. The resulting earth pressure is 

E = % -w - p -y = % X no X 12.2 Xi4 = 9394 lbs. 
The resultant R passes 9.3 ins. to the left of the center line a-a 
of the base. Using the formula deduced for footings with vary- 
ing pressures on the soil, page 240, f e = 7 — — {B =b 6 x), here 

W = 25,000 lbs. and is the vertical component of E and the 
weight of the wall, b = 1 foot, and the maximum pressure under 
the foundation is 

, 2^,000 / . 6 X 9-3\ c -m 

/. = -^ — i(io + ^ = 3670 lbs. 

1 X io 2 \ 12 / 

Since h±Jl X B = 25,000, /i = 1330 lbs. 

2 

An inspection of the wall will show that by increasing the 
width of the footing to 12 ft., as shown in Fig. 301, the resultant 



RETAINING WALLS 



289 



would be brought practically through the center of the base thus 
making the pressure uniform across it. 

To prevent sliding the coefficient of friction would have to be 
8100 -r- 25,000 = 0.324. As the coefficient of friction probably 
lies between 0.500 and 0.750 there is evidently ample margin of 
safety against failure in this way. 




8100* 



Fig. 300. 



Fig. 301. 



Reinforced-concrete Retaining Walls 

The use of reinforced concrete in retaining walls has led to 
some modification of the ordinary design. This results in con- 
siderable economy in walls exceeding 25 ft. in height. The com- 
mon section of such a wall is shown in Fig. 302. The vertical 
load upon the soil is the weight of the wall plus the weight of the 
soil carried directly by the wall, that is, the soil prism abed and 
the vertical component of the force E acting upon the back of 
the wall. The general lines of the reinforcement are indicated 
by the steel shown in Fig. 302 . The buttresses B are placed at 
intervals of from 8 to 10 ft. along the wall. Walls under 18 ft. 



290 



GRAPHICS AND STRUCTURAL DESIGN 





in 

6 



2 

CO 

6 



, 'i'li I 1, 11 ' . 1 'i ! 









JB.-4— 4—J---4. WU 

t-ft^r-t-4444 



S-^F-TF=|f =*=«"- IfjjPJ ft =ff 

!h4.Q!k^-d!v=!!---o li ^! |; 'aBct 



■ 11 11 ii 11 H H 11 11 11 1 „ H H 1, " . 

™ " 'I II II II II II II N l| || II 'I || II / 

' ' 11 11 « 1 !i n H " 11 11 11/ * 

I ; 11 11 11 11 1 n 11 n 11 11 ii 11 
(I , II II ll II II II II I II II II II 

II II II II II II J II J I J II II II II 




V<<M 



RETAINING WALLS 



291 



may have the buttresses omitted and the wall designed as a 
cantilever beam. 

Problem. — Design a reinforced-concrete wall for a height of 
25 ft. 

Assume the weight of the fill as 100 lbs. per cu. ft. The angle 
of repose of the material behind the wall can be taken as 30 de- 
grees. The maximum soil pressure is not to exceed 2 tons per 
sq. ft. The working unit stress in the steel is not to exceed 
12,000 lbs. per sq. in., while that in the concrete is not to be over 
500 lbs. per sq. in. 

The fill abed, Fig. 303, in the wall will act to prevent over- 
turning, so that we will first want the weight and the center of 
gravity of the wall, buttresses and prism abed of the fill. 



Section. 



Coping 

Face of wall 

Fillet 

Fillet 

Base 

Base 

Buttress 

Buttress 

Fill, prism over base . . . 
Fill, prism over buttress 

Total for 10 ft. length. 



Volume. 



Weight. 



25X140 = 

217.5X140 = 

7.5X140 = 

7.5X140 = 

280X140- 

15X140 = 

128.1X140 = 

22.9X140 = 

I759-5XIOO: 

160.0X100: 



3.500 
30,450 

1,050 

1,050 
39,200 

2,IOO 

17,934 

3,206 

175,950 

16,000 

290,440 



Arm. 


9-33 


9 


50 


10 


25 


10 


83 


7 


00 





75 


5 


55 


8 


67 


4 


50 


3 


00 



Moment. 



32,670 

289,275 

10,760 

H,372 

274,400 

i,575 

99.534 

27,785 

79L775 

48,000 

1,587,146 



Distance to center of gravity of wall, x = = 5.46 ft. 

290,440 

E = i(w-p-y) = i (100 X 1443 X 1443) = 10,400 lbs. 



For a length of 10 ft. this is 104,000 lbs. 

The diagram made to an enlarged scale of Fig. 303 gives the 
line of action of the resultant of E and the total weight, 290,440 
lbs., as passing through a point 35.8 ins. to the left of the vertical 
line through the center of gravity. The distribution of this 



292 GRAPHICS AND STRUCTURAL DESIGN 

pressure upon the foundation is given by the formula 

/.-^i(B±6*). 

B = 14, x = ^^ = 1.44 ft. 
12 

Sliding of the Wall. — The coefficient of friction between the 
wall and the soil will lie between 50 per cent and 75 per cent. 
The resistance to sliding, even assuming the coefficient at 50 per 
cent, would be 290,440 X 0.50 = 145,220 lbs. This, being con- 
siderably above the horizontal pressure on the wall, 104,000 lbs., 
indicates little possibility of failure by sliding. 

Reinforcements 

Face of the Wall. — The wall will be tied into the buttresses 

W -L* 

and the bending will be assumed as M = The reinf orce- 

12 

ments will be determined for sections down the wall and the 

bending moment will be calculated from the load taken from 

Fig. 304. As the pressure is assumed as varying proportionally 

to the depth, the distribution will be represented by a triangle. 

The area of this triangle equals 104,000 lbs., or 

104,000 = 0.5 X ac X ab or ab = (104,000 X 2) -f 25 = 8320 lbs. 

The load upon a strip 1 ft. wide and 10 ft. span is then given 
by an intercept in the triangle abc parallel to ab, the intercept 
being located the same distance from c that the center line of the 
section is from the top of the wall; thus the load on a strip 
12 ins. wide whose center is io ft. from the top is 3320 lbs. when 
scaled from Fig. 304. 

* Note. — If the negative moments are not to be carefully provided for 

u . ,, kn WL WL 

M should be taken or even — — • 

10 8 



RETAINING WALLS 293 

W • L 
The bending moment upon such a strip is M = = 3320 

X 10 X y| = 33,200 in. lbs. Using the approximate formula 

M 8 = %XAXf 8 Xd. 

Taking d = 10.5 ins. and/ 8 = 14,000 

. A 33,200 X 8 o • 

makes A = — = 0.258 sq. m. 

7 X 14,000 X 10.5 

§-in. bars would require a spacing of 

Area of J-in. X 12 0.106 X 12 A , 

^— = — z — - — , approximately 9 ins. 

^4 0.258 

If desired, this can be checked by the more accurate formulae. 

. • 0.258 

p — = 0.0020. 

10.5 X 12 

p • n = 0.0020 X 15 = 0.032. From the curves, page 202, 
j = 0.924; k -j = 0.231. 

Substituting these values in formulae (3) and (4), page 200, 

M.-AX/.Xj.i or /.-j^ 

- ^^,200 „ 

f 8 = o — = x 3,300 lbs. 

^ 0.258X0.924X10.5 °'° 

j r 2 ^ 2 X 33,200 to „ 

and / c = ; — : : — ~ = — ; = 217 lbs. 

J k-jXb-d 2 0.231 X 12 X 10.5 2 ' 

These values are satisfactory, it not being desired to make the 
wall thinner. The |-in. round bars will be spaced 8 ins. center 
to center in the section between 8 ft. and 10 ft. from the top of 
the wall. 

The reinforcements may be determined for the other sections 
in a similar way. The results of such calculations are given in 
the table. The calculated spacing is given in the brackets, while 
that to be used is given outside. 



294 



GRAPHICS AND STRUCTURAL DESIGN 



Depth of 

section, 

feet. 


Load, 
pounds. 


M W ' L 

M = • 

12 
inch pounds. 


Area of steel, 
square inches. 


Size of bars and spacing, 
inches. 


4 

7 

IO 

12 

14 
16 
18 

20 
22 


1400 
2400 

33 2 ° 
4100 
4800 
5400 
6lOO 
6750 
7400 


14,000 
24,000 
33,200 
41,000 
48,000 
54,000 
61,000 
67,500 
74,000 


0. 109 
O.187 
O.258 
0.318 

0-373 
O.420 

0-474 
0.524 
0-575 


¥0 spaced (21.6) 12" 
\"0 spaced (12.6) 12" 
\"0 spaced (9.1) 8" 
¥0 spaced ( 7.4) 7-5" 
¥0 spaced (6.3) 6" 
¥0 spaced (5.6) 5.5" 
¥0 spaced (4.96) 5-°" 
¥0 spaced ( 4.5) 4.5" 
\ "0 spaced (4.1) 4.0" 



The toe or base of wall to the left of the face will carry a load 

. t.. , 1 SSSo + 2600 ,, ,± 

given in Fig. 305 and equals "^^ = 2975 lbs. per sq. ft., 

2 

making a load per foot of wall of 2975 X 4 = 11,900 lbs. The 
distance of the center of gravity of this loading from the face of 
the wall is 



x _ 335o + 2 (2600) x 4§ 
3350 + 2600 3 



23 ins. 



The moment on a strip 12 ins. wide then is M = 11,900 X 25 = 
297,500 in. lbs. 
By the approximate formula M s = f X A X f 8 X d. 

. 297,500 X 8 

A = :LL±a = 1. 10 sq. ms. 

7 X 14,000 X 22 

The spacing for i-in. bars is 5 =8.6 ins. 

1.1 

Rear Portion of Base. — The load upon this section (see Figs. 
305 and 306) will be the difference between the pressure upon the 
soil and the weight of the fill vertically over this portion. This 
amounts to 23 ft. of soil and 2 ft. of concrete, and weighs 
2300 + 280 = 2600, approximately. The maximum loading is 
2600 — 800 = 1800 lbs. Estimating the reinforcement for maxi- 
mum loading the total load on a strip 12 ins. wide and 10 ft. 
long is 18,000 lbs. 

,, 18,000 X 10 X 12 . „ 

M = — • = 180,000 in. lbs. 



12 



RETAINING WALLS 295 

Placing rods 3 ins. above the bottom of the slab and using the 
approximate formula gives, 

8X1 8 X 180,000 



7 Xf s -d 7 X 14,000 X 21 



= 0.70 sq. m. 



m , . ,7. ^, .„ , 0.60 X 12 
The spacing of ^-m. bars will be = 10 ins., approx- 
imately. 

In assuming the bending moment at M = instead of — - — 

12 8 

it becomes necessary to place reinforcement in the outer flange 
over the supports; this would have to equal the reinforcement 
at the middle. Here short lengths of rods the same as the full- 
length reinforcements (see Fig. 306) will be placed over the sup- 
ports and spaced the same as the other rods. 

Counterforts. — In this design (see Fig. 302) these will resist 
the moment due to the thrust on the wall. The thrust on the 
wall from the top to the upper face of the footing is ^\— X 23 
= 87,975 lbs. 

The moment is 88,000 X — = 8,096,000 in. lbs. 

3 
The design can be made as a T beam, the wall serving as the 

flange. 

The distance from the face of the wall to the center of the 
reinforcement can be assumed as 108 ins. The ratio of flange 
thickness to depth of beam is ■£$$ = 0.11. From the curves for 
T beams, j = 0.95, and substituting in the equation 

M = A Xf 3 Xj-d, 

we have 

. M 8,006,000 , 

A = : — : = — — = 5.65 sq. ms. 

f 3 Xj-d 14,000 X 0.95 X 108 H 

This will require eight yf -in. rods. 

The number of rods can be reduced towards the top of the 
wall as the bending moment becomes less. 



296 



GRAPHICS AND STRUCTURAL DESIGN 



In a beam similar to the counterfort the flange force at any 
distance x from the top can be approximately expressed as 



height of wall whose flange force is F. 



(F • x 2 ) 
F x = - — -r- 1 , where h 
h 2 

This is not accurate, as j will vary with the depth of the beam, 
but may be approximated at t 9 q. The flange forces will be pro- 
portional to the required areas. The varying flange forces can 
then be represented by laying off a base line equal to the height 
of the wall and plotting ordinates according to the formula just 
stated. The curve being a parabola can be laid off as in Fig. 307. 

ca represents the side of the wall, here 25 ft., and cd represents 
the side extending to the top of the slab, de is the force in the 
steel 23 ft. from the top of the wall, gc represents the total 
length of diagonal steel reinforcements, the distance df measuring 
8 ft. from the wall. 

Eight rods taken in pairs will divide de into four equal parts 
and dh, hi, ij and je each represents the area of two of the rods. 
To find the lengths of the shortest rods 
produce jk perpendicularly until it 
cuts the curve, from k project a 
horizontal line until it cuts the diag- 
onal line in I. The lengths of these 
two rods will then be given by the 
length gl. As this curve gives the 
rate of change of flange force it also 
gives the horizontal shear which will 
be the difference between two adja- 
cent abscissas and the stirrups can be 
determined as shown for beams with 
uniform loads, page 218. The beam 
being so deep the concrete alone 
would probably supply ample strength 
for horizontal shear but in practice 
there are generally also metal stirrups inserted. 

Another type of wall is the can tile vered wall. In it the 




RETAINING WALLS 297 

buttresses or counterforts are omitted. This type is suited to 
lower walls than the design just carried out. The principal 
reinforcements are shown in Fig. 308. The calculation of the 
thicknesses and reinforcing metal can be made on the same 
general lines as the preceding problem. 

The vertical wall CD is treated as a cantilever and is assumed 
as carried by the base AB. 



CHAPTER XVIII 
BINS 

Pressures on Bins 

The methods used in obtaining the pressures upon retaining 
walls are also applicable to the pressures acting on the sides and 
bottoms of bins. 

Bins may serve as retainers of any material but the discussions 
here will be confined to such bins as are ordinarily used for coal, 
sand, refractory materials at steel mills, and similar materials. 
Bins at grain elevators are generally high compared with their 
width; it therefore follows that the friction of the material against 
the sides of such bins may greatly assist in relieving the lower 
portion of the bin, both sides and bottom, of this accumulative 
pressure. The theory previously given for retaining walls does 
not apply in this case as the assumed plane of rupture to produce 
maximum pressure will not cut the surface of the grain. Several 
formulae have been proposed for estimating these pressures; that 
of Janssen, taken from Hutte des Ingenieurs Taschenbuch, is 

T7 Rwf -|xA 

and L = V . k. 

L = lateral pressure of the grain, pounds per square foot. 
V = vertical pressure of the grain, pounds per square foot. 
/ = coefficient of friction of grain on the bin wall. 
w = weight of grain, pounds per cubic foot. 
R = hydraulic radius of horizontal section of bin. For circu- 
lar section R = . For other sections R = 

4 

area of section .„ ,. . . , 

All dimensions in feet. 



perimeter of section 

298 



BINS 299 

k = ratio of lateral to vertical pressure. 
e = base of Naperian system of logarithms. 
h = depth to bottom or point on side at which pressure is to 
be determined. 

For wheat, w = 50 lbs. per cu. ft.; k = 0.60; / = 0.40. 



Shallow Bins 

The bins now taken up will be shallow compared with those 
just considered. In these the theory of retaining walls will apply. 
These bins will commonly be made of, (a) Timber, (b) Steel, 
unlined, or lined with timber or concrete, (c) Reinforced concrete. 

The bin should be built or lining placed to avoid pockets that 
might hold materials indefinitely. This is particularly neces- 
sary in materials in which spontaneous combustion might occur. 
Timber lining should be tar coated on the sides and faces against 
the metal, when such treatment will not affect the material held. 

The inclined sides of unlined steel bins frequently have wear- 
ing strips, flats about 3 X § in., placed on about 12-in. centers 
and running at right angles to the direction of flow of the material, 
thus tending to retain a slight thickness of the contained material 
along the side or make the material slide upon itself rather than 
on the metal and so protect the metal from wear. 

Stresses in Bins 

Although the forces acting upon the bin sides may be esti- 
mated by formulae, the graphical methods used on retaining 
walls will be largely used in the following discussions. For those 
who prefer formulae those given by R. W. Dull in the Engineering 
News of July 21, 1904, are here given, with slight modification. 



3°° 



where 



GRAPHICS AND STRUCTURAL DESIGN 

wh 2 






cos 5 



2 w h 2 

2 



+ 8) sin 4> 



cos 5 



where 
If 




Fig. 309. Vertical wall, no surcharge. 



£ = ^X 



/ cos 4> y 



_ w& 2 / cos <£ V 

2 \I + «/ 



« = y s -Hi* 



+ 5) sin (cf> — a) 



cos 5 cos a 
ex = <f>, then « = o. 




Fig. 310. . Vertical wall, surcharged. 



BINS 



where 



E = 



N 



wh 2 
2 COS 5 

wh 2 _ 



( COS (f> \ 2 



„ = i / sin (<ft + 8) sin (0 + a) 
V cos 5 cos a 




Fig. 311. Vertical wall, surcharge negative. 



3P 1 



where 



= ^ 2 v r_cosj^_-0)_-|2 

2 cos 08 + 5) A L> + i)co S/ 3j ! 

wA 2 cos 5_ |~ cos 08 — 0) ~| 2 
5) X L(w+i)cos/3j ' 



N = 



2 cos (/3 + 



_ a / sin (0 + 8) sin <fr 
W ~ Vcos(5 +/3) cos/3 ' 




Fig. 312. Wall inclined outward. < 90 + 5. Surface horizontal. 



302 



GRAPHICS AND STRUCTURAL DESIGN 



where 



2cos(5+/3) A 



r cos (0 - /g) -i» 

L(» + i)cos|Sj : 



_ wff cos 5 r ,cos(</>-/9) ~] 2 

2 cos (5 +0) X L(» + i) COS/sJ ' 

_ * /sin (0 + 8) sin (0 — a) 
W ~ V cos (5 + 0) cos (fi-a)' 




Fig. 313. Wall inclined outward. 9 < 90 + 5. Surcharged. 
W = weight of triangle ABC = wA?tan ^ £ = VjF 2 _j_ ^ 



JS = — \ 



(«-!) 



tan/x 



tan/8 



tan 2 



(«-!) 



Q = E cos (/x - /3) and T = £ sin 0* - £)• 




Fig. 314. Wall inclined outward. > 90°+ 5. Surface level. 



BINS 



303 



„ . . cos a — Vcos 2 a — cos 2 <t> w wh 2 

Ei = cosa X , == X — L - 

cos a + V cos 2 o; — cos 2 2 

£1 acts parallel to line of fill. 

wh 2 sin (3 cos (a — 0) 



W = 



2 cos 2 /3 cos o; 



£ = VEi 2 + W 2 + 2 EiW sin a. 




Fig. 315. Wall inclined outward. 9 > 90 + 5. Surcharged. 



where 



Ni = tan 2 

wtan/3 



(«•-?) 



(tf 2 - #), 



w 



E 2 



(H 2 -h 2 ) -£ 2 sin5 2 , 



( cos <t> Y 



wh 2 

2 COS 62 



and 



— y/i 



sin (<f > + g 2 ) sin <fr 
cos 62 



£ = ViVx 2 + W 2 , 

2 £ 2 sin 1 



tan/3 - 



tan/z = 



w{H^-¥) 



tan 2 



(«-!) 



Q = £cos(m-/8); 

r = Esin(/i-/3). 

Note. — The quantity E 2 sin 5 2 is the 
friction on the vertical side and reduces 
the weight acting upon the side AB. 




Fig. 316. Hopper bin, fill level. 



3°4 



GRAPHICS AND STRUCTURAL DESIGN 



_ „ , cos a — Vcos 2 a — cos 2 </> v , w (hi 2 — h 2 ) 

Ei = cos a X , X : • 

cos a + V cos 2 a — cos 2 <f> 2 

m sin p cos («-g) . • _ ^ _ _ s . 

cos 2 cos a: 2 



£•> 



/ COS<A 

U + ij 



2 wfr 2 

2 COS §2 



where 



= i/ sin (0 + 52) sin (0 - a) 
» cos 6 2 cos a 



£ = VE1 2 + W 2 + 2 £iP^ sin a. 
£1 is parallel to the angle of the fill. 



Fig. 317. Hopper bin, surcharged, 
unsymmetrical. 




\n + 1 / 2 

, 4 /sin </> sin (<j> + a) 

where w = V 

t cos a: 



w = sin£cos( a -/3) x w _ + ^ _ ^ ^ _ ^.^ 
cos 2 /? cos a 2 

Q = E cos (p — 0) and 

r = £sin(M-/3). 



Fig. 318. Hopper bin, surcharged, sym- 
metrical. 




BINS 



3°5 



The following coefficients of friction are given by Mr. Dull in 
the article just referred to. 

COEFFICIENTS OF FRICTION BETWEEN MATERIALS 



Material. 


<t> Degrees. 


1 Degrees. 


Weight per cu. 
ft., pounds. 


Bituminous coal 


35 

27 

34 

40 


18 
16 
18 
31 


50 
52 
90 
40 


Anthracite coal 


Sand 

Ashes 







Where the material runs on concrete 4> l may be assumed equal 
to <f>. 

<p = coefficient of friction of the material upon itself. 
1 = coefficient of friction of the material upon steel. 




1 Column Load 
I ,150900^ 
Fig. 319. Fig. 320. 

Graphical Determination of Forces Acting on Bin 

Bin not Surcharged. — Assume Fig. 319 to be a bin for bitumi- 
nous coal. The coal weighs 50 lbs. per cu. ft. <£ = 35 degrees. 
The natural sine of 35 degrees is 0.574. 



306 GRAPHICS AND STRUCTURAL DESIGN 

The pressure against the side AB and the plane CI may be 
found as was done for retaining walls with Rankine's formula. 
OnAB, 

El . l w #l=m± = 5£ Xlo2x L^574 _ 6 lbs 

2 i+sm<|> 2 i -+- 0.574 

On/C, 

^ 1 79 i— sin</> 50 w 9 v 0.426 „ 

E2 = - ^ 2 r~^ = — X 20 2 X — *— =2700 lbs. 

2 1 + sm<£ 2 J -574 

The side JBC is acted on by the weight of the superimposed coal 
and the horizontal pressure E^. It is easier to consider the 
triangle of coal CJI, and then having found the pressure on CJ 
the pressure on the section CB may be scaled from the diagram. 
The weight of the triangle of coal CIJ is W = \ X 20 2 X 50 = 
10,000 lbs. 
This resultant force acts at the center of gravity of this tri- 

angle or - JI from CI. Eq acts a distance CK = — from the 

3 3 

bottom of the bin. Combining E2 and W gives the resultant 
R = 10,300 lbs. This force resolved parallel and at right angles 
to the side BC gives T = 5200 lbs. and N = 8900 lbs. The 
normal pressure of 8900 lbs. is distributed along the line JC as 
the intercepts in the triangle LJC. Putting the area of the 
triangle LJC = 8900 lbs. and solving for the side LC gives LC = 
630 lbs., which represents the pressure per square foot at the 
point C. The pressure at B is found, by scaling the intercept 
BM , to be 310 lbs. 

The total pressure on the side BC is the area of the trapezoid 
BCLM and equals 6580 lbs. The resultant of this pressure acts 
through the center of gravity of the trapezoid. The forces 
acting upon the points of the bin with supports every 12 ft. are 
given in Fig. 320, and are the force a,tA= (-|--) X 12 = 2700 lbs. 
The horizontal force atjB = (|) X675 X 12 = 5400 lbs. 

The force at B normal to the side BC must be estimated as 



BINS 307 

follows. The total normal pressure on BC is 6580 lbs. CQ = 
6.5 ft. CB = 14.1 ft. The resultant at B is, therefore, 

(6580 X 6.5) „ 

— — = 3040 lbs. 

14.1 

The total on 12 ft. of bin is 3040 X 12 = 36,480 lbs. The balance 
of the load represented by the trapezoid MBCL acts at C and is 
(6580 X 12) — 36,480 = 42,480 lbs. 

The vertical pressure at C is CD X DH XwX 12 = 5 X20X 
50 X 12 = 60,000 lbs. 

The load on the columns due to the fill will be the total weight 
of the coal in the bin. The area of the bin section is 500 sq. ft., 
making the load in 12 ft. 500 X 12 X 50 = 300,000 lbs. 

The trussed bracing will be carried to the points A, B, C, E, 
F and G of the bin and taking these forces together with the 
total column load the several forces acting in the frame may be 
determined. 

Surcharged Bins. — A bin similar to the last one but sur- 
charged (Fig. 321) may have the forces acting upon it analyzed 
in much the same way. First find the pressure upon the side 
AB\ this is done graphically in Fig. 322. The diagram gives 
p = 8 ft. and y = 8.5 ft., from which E 1 =^XwXpXy = 
i X 50 X 8 X 8.5 = 1700 lbs. The normal pressure is 1620 lbs. 

Now finding the pressure upon IH, the axis of the bin, by using 
the graphical method and assuming 4> l = o, we have p = y = 
16 ft. 

£2 = 0.5 X w X p X y = 0.5 X 50 X 16 X 16 = 6400 lbs. 

The center of gravity of the triangle IJH is at M and the 
weight of a prism of coal 1 ft. high with the base IJH is 

(35.4 X 20. 7 X 5 0) lu 

* s2j2 - a: ■M- = 18,300 lbs. 

2 

The resultant of E2 and W is R = 19,300 lbs. This is assumed 
as varying uniformly along the face //. The pressure at / is 

x _ (19,300 x 2) = I45olbs- = /Ar . 
26.5 



3°8 



GRAPHICS AND STRUCTURAL DESIGN 



The total pressure on the side BC is that represented by the 
trapezoid QBCS whose altitude scales 12.8 ft.; the area then is 

410 -f 1 1 00 
2 

lH 



X 12.8 = 9660 lbs. 




Fig. 321. 

This acts through the center of gravity of the trapezoid 

which has been located graphically. 
The normal pressure on the side 
BC may now be found by laying 
off 9660 lbs. through the point 
and resolving it into its com- 
ponents parallel with and at right 
angles to BC. The load upon 
one-half of the base CD is 7210 lbs. 
The forces, due to the fill, acting at 
the points of the bin may be esti- 
mated as in the preceding problem and are given in Fig. 3 23 . The 
distance between supports has been taken at 1 2 ft. The columns 




Fig. 323. 



BINS 



309 



will carry the total weight of the coal in the bin and the weight 
of the bin. The stresses in the internal framework of the bin 
may be calculated by treating these external loadings as is done 
for any other truss. The stress diagram may be drawn or the 
forces calculated by any of the methods previously given. 



Suspension Bunkers 

If the side plates of the ordinary hopper bunker are permitted to 
bulge but slightly they must be supported at frequent intervals by 
beams or other structural shapes. This added weight is avoided 
in the use of suspension bunkers. These bunkers are patented 
and are designed with the idea that the sides resist tension only. 
To fulfil this condition theoretically the sides would assume a 
shape peculiar to each possible loading. This could be illustrated 
by a model bin whose sides were a purely tension piece like muslin 
or duck. However, an actual bin differs greatly from the above 
illustration and any deflection due to bending in the sides when 

the bin is not loaded as theoretically 
assumed is ordinarily not sufficient 
to cause trouble. The lined bins 
will be stiffer than the unlined ones. 
The theoretical curve of a bin for 
any assumed loading may be deter- 
mined as follows. 
In Fig. 324, given 
the desired span 
and sag, the load 
in the left-hand side 
of the bin will not 
vary much from a 
The weight of the material in the portion 
will act through the center of gravity of 




SUSPENSION 
BUNKERS. 

Fig. 324. 




Fig. 325. 



triangle ABC. 

ABC of the bin 

this triangle ABC and lies in the vertical fine DG or W, GF 

being — ■• 
3 



310 GRAPHICS AND STRUCTURAL DESIGN 

Considering the side of the bin BE the force acting at B and 
that at E must hold W in equilibrium; hence they must pass 
through a common point D. 

The line of action of W is known as is also the line of action of 
the force acting at E which must be horizontal if the bin is sym- 
metrical and symmetrically loaded, both of which conditions have 
been assumed. If then a horizontal line is drawn through E it 
cuts W in D and the line of action of the force acting at B must 
lie in the line BD passing through the point D. The loading 
being assumed as varying as the intercepts of a triangle we may 
represent this loading by the triangle FBC, and having divided 
it into sections of equal widths the weight of each section will 
be proportional to its center ordinate; these have been shown 
dotted. 

These lengths from 1-2 to 5-6 have been laid or! on the load 
line of the force polygon, Fig. 325. The lines of action of the 
forces acting at B and E being known the pole O is readily found 
by drawing Oi parallel to DB and 06 parallel to BE. The other 
strings may now be drawn in the force polygon. If we complete 
the equilibrium polygon in Fig. 324 the broken- line side will 
approximate the theoretical side of the bunker. On the right- 
hand side the line of the bunker has been drawn as it is com- 
monly made in practice. The portion HI is straight while the 
lower part HE is an arc of a circle. The sides of this bin 
approximate an equilateral triangle. The sag is about six-tenths 
of the span. 

For a bin which is nearly an equilateral triangle the following 
figures will be approximately true. 

Area of the bin section, fill level, = 0.40 S 2 . 

Area of the bin section, triangular surcharge, = 0.57 S 2 . 

Length of plate, no allowance for laps, = 19.8 5 (inches). 

Force in plate per foot of length at B or / = 0.60 W (pounds). 

Force in plate per foot of length at £ = 0.30 W (pounds). 

Here S = the span in feet. 

W = the load in the bin for one foot of its length, pounds. 



BINS 



311 




Fig. 326. 



In the case of the suspension bunker with vertical sides above 

the points of suspension, B and / of 
the bin, Fig. 326, the cross section 
of the bunker will be increased by 
the area BCIJKL equal to S X h, 
and the load will be increased pro- 
portionally. In this case the weight 
of the material in one-half of the bin 
will act through the center of gravity 
of the section KLBDEK. This will 
be a greater distance from the cen- 
ter line EC than was the case in the 
preceding problem, and will make 

the bin somewhat wider below BI than at corresponding points 

in the former bin. 

W 
In either problem if — is known the forces acting at B and E 
2 

may be found by drawing the triangle B 1 E l D 1 at the side of the 

bin as in Fig. 326. 

One form of Suspension Bunker with column supports for 

boiler rooms is shown in Fig. 327. This bunker is intended to 

hold 6600 lbs. per ft. of length. Assume coal as weighing 50 lbs. 

per cu. ft., and the sag of the bunker as 60 per cent of the span. 

The bunker will be assumed as carrying the required load when 

surcharged, and the angle of repose of the coal will be taken 30 

degrees. Determine the bin area by using the formula, A = 

0.57 S 2 . The required area is ^-frr 2 " = T 3 2 sc l- ^- From this 

132 = 0.57 S 2 and S = 15.3 ft. If the bin is made 16 ft. wide 

the sag will be approximately 10 ft. Spacing the columns 20 ft. 

center to center makes the load of coal upon each column 

= 66,000 lbs. Adding 4000 lbs. to this to cover 

2 

the weight of metal, exclusive of columns and under bracing, 

makes a total of 70,000 lbs. at each column. This load laid off 

in Fig. 328 gives the stress in the side of the bunker per foot of 



312 



GRAPHICS AND STRUCTURAL DESIGN 



length as — l = 4000 lbs. The unit stress per square inch 

20 

in a plate - in. thick is -, — r = 1330 lbs. 

4 (0.25 X 12) 

The main columns are subjected to a bending moment of 
80,000 [(16 — 6) — 6] = 320,000 in. lbs. 

Fig. 328. 




Fig. 329. 
Trying two 12-in. channels weighing 20.5 lbs. per ft., we have 

2 2 O OOO 

Fiber stress due to the bending moment is , ' — tt = 7,400 lbs. per sq. in. 



Fiber stress due to direct loading is 



(2 X 21.6) 

80,000 
(2 X 6.02) 

Combined stress 



= 6,650 lbs. per sq. in. 



= 14,050 lbs. per sq. in. 

The direct load upon the column being 6650 lbs. per sq. in. the 



BINS 313 

radius of gyration for an assumed length of 1 1 ft. may be found 
from the straight-line formula, 

/ = 16,000 — 7o(-)= 16,000 — 70 [-^-), 

70 X 132 

or r = -r—1 -f- — r = 0.99. 

(16,000— 6650) 

Two 12-inch channels are to be used back to back. The 
radius of gyration of a 12-inch channel referred to an axis at the 
back of the web is 



r = Jl±A!*L = y/ 3-9° + (6.02 X 0.70 2 ) _ lo6 
* A * 6.02 

It therefore follows that two 12-in. channels at 20.5 lbs. per ft. 

either riveted back to back or separated any distance will carry 

the load of 80,000 lbs. 

The horizontal strut under the bin carries a load of 40,000 lbs. 

Its length is 84 ins. Trying a 10-in. channel at 20 lbs. per ft., 

whose radius of gyration about an axis parallel to the web 
7 o . 

is 0.70, we find- = — — = 120. The allowable stress is / = 
' ' r 0.70 J 

16,000 — (70 X 120) = 7600 lbs. per sq. in. The required area 

then is-*— - = 5.26 sq. ins. The combined bending and direct 

7600 D H & 

stress on this piece will exceed that allowed by paragraph 93 of 
the specifications. Considering the design and character of the 
connections to the columns it will probably be ample; if pre- 
ferred a four angle latticed strut will meet the requirements but 
its fabrication would cost more. The angles could be 2J" X 
2 \" X \" . The strut across the top of the bin carries the con- 
veyor, its truss and the conveyor loading and in addition to 
these the direct compression due to the bin loading. Assuming 
the load equivalent to a central load of 3500 lbs. the bending is 

W >l 12 

M = = 3500 X 16 X — = 168,000 in. lbs. Trying a 10- 

4 4 



314 GRAPHICS AND STRUCTURAL DESIGN 

in. I beam at 25 lbs. per ft., with an- value of 24.6 and an area 

e 

of 7.34 sq. ins., we have 

Fiber stress due to bending moment, - = 6800 lbs. per 

24.6 

sq. in. 

Fiber stress due to direct loading, — = 5450 lbs. per sq. in. 

The combined fiber stress is 6800 + 5450 = 12,250 lbs. per sq. 
in. Considering that this beam is held laterally at its center 
by the conveyor truss this fiber stress would seem permissible. 
The braces occur every 10 ft. 

The side of the bin between columns is carried by a girder. 
The total load is 80,000 lbs. or 4000 lbs. per ft. The bending 
moment then is 

Wl 12 

M = — = 80,000 X 20 X — = 2,400,000 in. lbs. 
8 8 

Assuming the distance back to back of flange angles as 3 ft., 
the distance between the centers of gravity of the two flanges 
will be about 34 ins. 

M = A Xf X h = 2,400,000 = A X 16,000 X 34 
or A = 4.42 sq. ins. 

Two 5 X 3! X i 5 6-in. angles having a gross area of 5.12 sq. ins. 
and a net section, allowing for one rivet, of 4.58 sq. ins. furnish 
just a little more than the area required. No allowance has here 
been made for the web acting as flange. This gives additional 
security. 

Figure 329 shows a bin designed for bituminous coal. The 
weight of the coal was taken 50 lbs. per cu. ft., the natural 
slope was taken 30 degrees, while the angle of friction between 
the sides and the coal was assumed as 18 degrees. The diagram 
gives p = 16 ft. and ^ = 17 ft. 

E = % Xw-p-y = % X 50 X 16X17 = 6800 lbs. 

The horizontal component of this force is 6500 lbs. The hori- 



BINS 315 

zontal pressure per square foot on the sides of the bin at the 
bottom is x = 6500 X 3 2 o = 433 ^ DS - 

The sides are supported by I beams spaced 6 ft. center to 
center and horizontal rods are run across the bin at 6-ft. intervals 
and placed 10 ft. above the bottom of the bin. The stresses were 
determined graphically as outlined for fixed and continuous 
beams. The force in the tie rods was 28,800 lbs. and the maxi- 
mum bending moment on the I beams was found to be 570,000 
in. lbs. Bolts i| ins. in diameter, having an area of 2.05 sq. ins. 

at the root of the threads, would have a unit stress of — l = 

2.05 

14,000 lbs. per sq. in. and are satisfactory. 
Allowing 20,000 lbs. per sq. in. fiber stress in the beam would 

/ , , I M 570,000 r™ . , , 

require an - value of - = — = — =28.5. This would re- 

e e ] 20,000 

quire a 10-in. I beam at 35 lbs. per ft. or a 12-in. I beam at 

31.5 lbs. per ft. This bin was tied across the top by the roof 

framing and was braced transversely at intervals against wind 

pressure. 

Bin Design 

Figure 330 illustrates a bin for coal storage over gas producers. 
The upper portion of the bin was a square of 13 ft. side. The 
lower part was a curve with a 9-ft. sag. The coal is assumed as 
weighing 50 lbs. per cu. ft. and its angle of repose has been taken 
35 degrees. 

The angle of friction between the coal and the sides of the 
bin has been taken 18 degrees. Owing to the large amount of 
coal carried in the square portion of the bin the loading cannot 
be represented as a triangle; therefore an approximate outline 
of the bin has been assumed in Fig. 330 and the actual curve 
estimated from this. The left side of the bin has been divided 
into four strips of equal widths; the weight in each strip per foot 
of length of bin will be proportional to the middle ordinate, 
shown in full fines for each section. In the force polygon, Fig. 
331, the forces may be represented by these middle ordinates or 



316 



GRAPHICS AND STRUCTURAL DESIGN 



by a constant percentage of them. In the figure the scale has 
been taken as one-fourth of these lines. The forces A 1 B 1 , IPC 1 , 
etc., have been laid off on the right-hand side of the bin equal to 
and symmetrically placed with those on the left-hand side. These 
forces have been laid off on the load line of the force polygon, 
Fig. 331, and the vector diagram, on its left-hand side, has been 





used to locate the center of gravity of the load in the right-hand 
side of the bin. The center of gravity is found in the inter- 
section of the lines 1 and 5 in Fig. 330. R, the total weight of 
the coal on this side of the bin, acts through this point of inter- 
section. The force in the extreme low point of the bin will be 
horizontal, and this together with the force in the bin side acting 
through c must hold R in equilibrium, hence these three forces 
must pass through a common point b ; this gives the direction of 
be, which is the theoretical direction of the side of the bin at the 
point c. 

The lines for the other side of the bin have been made similar 
to those for the right-hand side. The vector polygon drawn on 
the right-hand side of the load line, Fig. 331, has been used in 



BINS 



317 



drawing the dotted lines in Fig. 330, thus showing the agreement 
between the actual and the theoretical lines of the bin. 

Before the stresses can be read in Fig. 331, the scale must be 
found. 

The mean ordinate in one side of the bin is 21 ft. The bin 
being 13 ft. wide makes the volume of the bin section for a length 
of 1 ft. = 21 X 13 = 273 cu. ft. This gives 273 X 50 = 13,650 
lbs. of coal per ft. of bin. The load line AE = 6825 lbs. This 
makes the scale ^- 8 1 2 -^- = 1700 lbs. per in. (approximately). 

The force AO 1 = 4.3 X 1700 = 7300 lbs. 

The force EO 1 = 1.6 X 1700 = 2720 lbs. 

The bin plates will be made yg in. in thickness to allow for 
wear and deterioration, and will have 3 X |-in. wearing strips 
spaced about 12 ins. center to center on the curved portion of 
the bin sheets and running the length of the bin. The distance 
center to center of columns will be assumed as 18 ft. Trans- 
verse braces will be placed at the columns and every 9 ft. 





X \ s v / " 11 




Tl 


\. #=io.5\ jy^ 




.1 




3^y\ 


*f 


1 

i 


' N^.2700* 


\ 



Fig. 332. 




The pressure against the vertical sides of the bin is determined 
in Fig. 332. 

E = ±Xw-p-y = ^X 50 X 10.5 X n = 2890 lbs. 

It should be noted that for so narrow a bin this does not give 
the accurate pressure but it is the limiting pressure to which 
this force on the side of the bin could extend. A closer approxi- 
mation might have been made by assuming a bin, say, 16 ft. 
high instead of 13 ft. but with a horizontal fill. The horizontal 



318 GRAPHICS AND STRUCTURAL DESIGN 

component of this force of 2890 lbs. is 2700 lbs. Two- thirds of 
this, or 1800 lbs., act at each of the lower points e and c, while the 
other third, 900 lbs., acts at each of the upper points/ and g. 
These forces act in the opposite direction to the pull from the 
suspended portion of the bin. 

Had the bin been level, substitution in the formula for the 
pressure on the side of the bin would have given 1060 lbs. or 700 
lbs. at the points e and c; the net pressure here then would be 
2700 — 700 = 2000 lbs. per ft. of length of bin. 

The compression in the lower member of the brace due to the 
total load between the columns is 2000 X 18 = 36,000 lbs. If 

the value of -in these braces is to be limited to 120 then r = 13 
r 

X X27 = I -3> an d this requires two 6 X 3§-in. angles having 
their long legs placed back to back. Before the weight is selected 
the possibility of additional loading from wind should be con- 
sidered. 

In exposed places this might reach 30 lbs. per sq. ft. of vertical 
projection. Along the bin at the point e it would equal (6.5 + 9) 
X 30 = 465. This makes the total load at the column (465 X 
18) + 36,000 = 44,370 lbs. 

The unit stress on the angle is 

/ = 16,000 — (70 X -)= 7600 lbs. per sq. in. 

The area of the angles is ' = 5.83 sq. ins. 

7600 

The minimum angles should therefore prove ample, unless 
larger angles are used to provide for corrosion and possible injury 
from the coal passing over them. 

The channel at e is acted upon by the wind and the pull from 
the side of the bin. Considering first the wind load, it will act as 
a uniform load on the span of 18 ft. Its deflection under this 
load will put a central load on the channel at c. If P is the load 
transferred to the channel at c from the channel at e, the de- 



BINS 319 

flection due to the wind load W is -*- X (— — -), while the de- 

384 \E'II 

1 fP • / 3 \ 
flection due to the central load is — ■ X ( — — r). Since the 

48 VE-// 

deflections of the two channels are equal, 

5 Wl s _ 1 PI 3 _ 1 P/ 3 

384 £/ 48 £/ 48 £/ 

and 5i^ = j? or P= HP TF= ^ 

384 24 384 

The maximum bending moment may now be found by adding 
algebraically the bending moments due to these several loads. 
Both sides of the bin should be tried although here the maximum 
bending occurs on the windward side. The moment due to the 
wind load of 465 lbs. per ft. is 

M = W - - = 465 X 18 X 18 X V 2 = 226,000 in. lbs. 
8 

The bending moment due to the central load is 

M = P • - = 0.31 X 465 X 18 X 18 X V = 140,000 in. lbs. 
4 

The bending moment due to the uniform load of 2000 lbs. per ft. 

on the 9-ft. span, that is, between the braces, is 

M = W - - = 2000 X 9 X 9 X V 2 = 243,000 in. lbs. 
8 

The resultant bending moments are given by Fig. 333, and are 
the intercepts between the curve abc and the lower curves aedfc. 
The curve abc gives the moments due to the wind load acting 
on the 18-ft. span and measured from the base line ac. The 
moments due to the central load on the 18-ft. span are then 
deducted by laying off the triangle adc. The bending moments 
due to the uniform load on the 9-ft. span are then plotted below 
ad and dc, thus giving the algebraic sum of these bending mo- 
ments. The maximum moment scales 340,000 in. lbs. and would 

require a channel having an - value of - = — 7 = ^~ = 21.^, 

H & e e f 16,000 °' 



320 GRAPHICS AND STRUCTURAL DESIGN 

which calls for a 12-in. channel at 20.5 lbs. per ft. The fiber 

stress in this channel would then be 24— > = 15,900 lbs. per 

sq. in. 

In a similar way the maximum bending on the upper channels 
is found to be approximately 139,000 in. lbs. and requires a 

section modulus of -^ — =8.7. 
16,000 

The sides act as girders to transfer the loads to the columns. 

The uniform load due to fill is 13,650 lbs. per ft., and assuming 

the weight of the bin as approximately 1000 lbs. per ft. makes 

the total load per foot 14,650 lbs. The load per foot on each 

. , . 14,560 „ 

girder is — — — =7325 lbs. 
2 

The bending moment on the girder is 

M = E_l1 = 7325 x l8 x i_8. = 2g6)66o ft. lbs. 
8 

If the girder is assumed as 13 ft. deep the flange force is 

296,660 ,, 

- 2 - 1 = 22,800 lbs. 

13 
When it is considered that in the lower channel the splice plate 
to which the curved portion of the bin is attached assists in 
carrying this stress it will be seen that this force of 22,800 lbs. 
will add very little to the fiber stress in the 12-in. channel. 

If 10-in. channels at 15 lbs. per ft. are tried for the upper 
channels the fiber stress due to the bending is 

, M-e 1^0,000 ,, 

/ = — — = -* 3LZ2 — = 10,400 lbs. per sq. m. 

I i3-4 

The fiber stress due to the flange force of 22,800 lbs. is 



22,800 



5100. 



4.46 

The combined fiber stress is 10,400 -f 5100 = 15,500 lbs. per 
sq. in. 



BINS 



321 



The stresses may be increased by wind acting on the ends 
of the bin or building and also by thrust from the traveling 
crane, or the forces may be provided for by additional bracing. 

The stiffeners on the sides ef and cq of the bin carry a triangular 
load of 2700 lbs. per ft. of length of the bin; hence if the stiffeners 
are spaced 2 ft. 3 ins. center to center, their total load is 2700 X 
2.25 = 6070 lbs. The moment on a beam due to such a load is 
M = 0.128 XP -l; ^ = 0.128X6070X13X12 = 121,000 in. lbs. 

The section modulus required is 

/ M 



121,000 



= 7.6, 



e f 16,000 
calling for 8-in. channels at nj lbs. per ft. 










-^ 




-^ 


" """ 










1 

© 

1 "s i 



















! 1 














^\ v 18"Channel @ 20}£ it per foot ^^ 




X 




• 














■ 


■ 



Fig. 334. 



Fig. 335. 



The maximum stress in the jg -in. plates along the side of the 
bin will occur at e. Since the load of 2700 lbs. is assumed as 
distributed as a triangle of pressure, it follows that if p e is the 

pressure per square foot at e, then 2700 = and p e = 415 lbs. 

2 

per sq. ft. 



322 GRAPHICS AND STRUCTURAL DESIGN 

The pressure per square inch is \\^ = 2.88 lbs. 

1 a 2 b 2 p 
The maximum fiber stress is/ m = - cf> — — X V 

2 a 2 + b 2 t 2 

x 1 v. v. 2 7 2 X 156 2 w 2.88 „ 

/ m = - X 1.0 X -V: — Vo X ; = 10,600 lbs. per sq. in. 

2 27 2 + 156 2 0.31 2 

A cross section and end view of this bin is shown in Fig. 334, 
while the elevation is given in Fig. 335. The diagonal bracing 
shown in the half section has not been drawn in the elevation. 
It would be located at the columns and at the center line midway 
between them. In general only the bin proper has been shown, 
as the design of the wind bracing and columns would have been 
complicated by the character of the building and it was not 
desired to consider this design here. 



CHAPTER XIX 
SHOP FLOORS 

The purpose of a floor is primarily to carry such loads as men, 
machines and materials. In addition they should preferably 
be poor conductors of heat as this adds greatly to the comfort 
of men standing on them. 

They should also be durable, resisting wear from the traveling 
of men and the transportation of goods upon them. They should 
furthermore be impervious to moisture. In special cases addi- 
tional qualities may be desirable; thus in many floors it is essential 
that they shall be nearly dustless, while in other cases a somewhat 
soft surface may be desired. 

Earth and cinder make the cheapest floor. Owing to dust they 
are only adapted to places like forge-shops, where hot metals 
placed upon them or heavy pieces dropped on them would only 
injure other floors. The soil below the floor level should be 
drained if damp and the floor clay or cinder then placed in layers 
upon a rolled bed; each layer should be thoroughly tamped or 
rolled. Where the moisture is excessive this floor may be made 
impervious by a layer of tar concrete placed from 1.5 to 2 ft. 
below the floor surface. 

Cement Concrete Floors. — In a floor of this type it is necessary 
to have a substantial foundation; its depth will vary with the 
character of the soil from 1 to 2 ft. The soil having been ex- 
cavated to the desired depth the surface should be well tamped 
or rolled. A layer of broken stone, cinder or gravel 6 ins. to 
10 ins. thick is placed upon this and thoroughly .tamped. In 
some floors a layer of from 3 ins. to 6 ins. of finely crushed stone 
is placed upon this and well rammed, this being in turn covered 
with 2 ins. to 4 ins. of concrete upon which a wearing coat of 

323 




324 GRAPHICS AND STRUCTURAL DESIGN 

cement mortar from \ in. to i in. thick is placed. In other 
designs the concrete is placed upon the first foundation layer. 
This concrete is 3 ins. to 4 ins. thick and upon it a wearing coat 
of 1 in. of cement mortar is placed. 

The concrete is usually a mixture of 1 part cement, 2 to 3 parts 
of sand, and 5 to 6 parts of stone or gravel. The cement-mortar 
is 1 part cement to 2 parts of sand or may be richer in cement. 
Where the soil affords a sufficiently good foundation 5 ins. to 
6 ins. of concrete placed immediately upon it may prove satis- 
factory. 

Tar Concrete Floors. — Where a wooden wearing surface is 
desired such surface may be laid upon the concrete base just 

.Hemlock ,/lH Maple 

|9 *:-■. " l^'Sand 

; ; ■o. ■_ •. ■ ; ^.-. .; -. •-.-. -. -if •. v> -. : : • :o;-.-. - -_-.:< 

Fig. 336. 

described, in which case the surface of the concrete under the 
wood should be coated with coal tar or asphalt; otherwise the 
wood deteriorates from dry rot. 

Another plan (Fig. 336) makes the foundation of a coal tar 
or an asphalt concrete. After the top soil is removed the surface 
is leveled and rolled and a layer of from 4 ins. to 6 ins. of broken 
stone or gravel thoroughly mixed with tar is placed and well 
rolled. A i-in. layer of sand thoroughly saturated with tar is 
put on this concrete and carefully rolled. While the tar is still 
hot hemlock planking is put on the sand and pushed into place. 
The wearing surface, which is preferably maple strips, tongued 
and grooved, is placed at right angles to and upon this hemlock. 
In some cases the foundation is made like that described for a 
concrete floor. The design may also be modified by using hem- 
lock sleepers instead of the planking. 



SHOP FLOORS 325 

In this case the hemlock strips are placed on about 3-ft. to 
4-f t. centers, being imbedded in the sand and tar, and the tongued 
and grooved maple flooring is then nailed to the sleepers. 

Wooden Block Floors. — Wooden blocks set on end make an 
excellent floor. These blocks are about 4 ins. to 5 ins. thick and 
may be either cedar, maple, beech, oak or pine. When not 
treated with a preservative they should be set in coal tar. Also 
when not treated the blocks should be set with spaces between 
them to allow for swelling due to any moisture they may absorb. 
These spaces between the blocks should be rilled with tar and 
sand. These blocks may be placed upon the tar-concrete 
foundation just described; see Fig. 336. 

Asphalt Floors. — Asphalt makes a satisfactory floor in many 
places, but it is not suited to machine shops or places where much 
oil is likely to reach it. Asphalt gives a soft surface, is water- 
proof, dustless and a poor conductor of heat. It therefore makes 
a very comfortable floor for the men to stand upon. It is 
easily dented but the dents will gradually smooth out owing to 
the tendency of the asphalt to flow under pressure. 

The foundation is similar to that for a concrete wearing surface. 
The asphalt is generally made an inch thick. A wash of molten 
tar and asphalt should be put on the surface of the concrete 
before placing the asphalt. Where asphalt is placed upon a con- 
crete slab it forms an additional dead load and. should not be 
assumed as resisting any of the compression in the slab. 

Brick Floors. — Brick floors are specially adapted to places 
where a floor is liable to local injury, as no other floor is so readily 
repaired, it being possible to remove and replace a few bricks 
without disturbing a large section of the floor. It has there- 
fore found favor in railroad buildings such as round houses. For 
brick paving the foundation consists of from 4 to 6 ins. of cinder 
placed in layers and tamped or rolled upon the excavated surface 
which had also been previously compacted. A layer of sand 
from 1 to 2 ins. thick is placed and rolled upon the cinder and 
then the brick are laid upon the sand. The bricks should be 



326 GRAPHICS AND STRUCTURAL DESIGN 

hard, homogeneous and impervious to water. They should be 
set on edge and joints should be broken by a lap of at least 
3 ins. None but whole bricks should be used except when nec- 
essary as fillers. After laying, sand should be thoroughly worked 
into the spaces between the bricks. 

To make this floor waterproof the brick may be grouted with 
tar and pitch, after which the sand may be worked into the 
spaces as before. 

Wooden Floors. — These are commonly made by embedding 
stringers, timber strips, in a foundation similar to those described 
for concrete floors and then nailing the wearing surface of the 
floor on the stringers. 

One wooden floor had a foundation of 8 ins. of cinders, the 
stringers were spaced 3 ft. centers and 3 -in. planking was placed 
upon these. 

The stringers and under surface of the planks were coated 
with lime to preserve them. Present practice prefers tar to 
lime to preserve wood. In other cases the foundation is a layer 
of concrete in which the stringers are embedded ; the under sur- 
face of the wood should be coated with tar as before. 

Ground Floors in General. — Where much water is used on 
floors they should be given sufficient pitch to properly drain the 
water and suitable drains or gutters must be provided. 

Where heavy trucking is done tracks must be laid. These 
may be either tracks sunk in the floor or flat iron plates placed 
in the floor level with the wearing surface of the floor. The 
former method is objectionable on account of breaking into the 
surface of the floor. The flat plates should be corrugated to 
prevent their becoming too smooth. 

Upper Floors 

Floors above ground level usually consist of two parts, one 
affording the necessary strength to sustain the floor loads, the 
other supplying the wearing face. The first part may be wood, 
steel, concrete, brick or tile, or may be a combination of some 



SHOP FLOORS 



3 2 7 



of these. The wearing surface may be any of the wearing 
materials used upon the ground floors. 

Wooden Floors. — A wooden floor is commonly carried upon 
wooden joists or steel floor beams. The depth of the floor 
timbers furnishing the strength will vary with the distance 

>^1 Tongued and Grooved Maple 




Fig. 337. 

center to center of the joists or floor beams and with the floor 
loads and will range from 2 or 3-in. planking as in Fig. 337 to 
2 X 8-in. timbers placed on edge and nailed together as in Fig. 
338. These planks should be continuous over at least 2 supports. 
The joints should be staggered, that is not broken continuously 



S*\ Maple 




STEEL FLOOR BEAM 



Fig. 338. 

on the same line. The wearing surface will usually be 1 to 
1.5-in. maple or other hardwood flooring. Frequently 3 thick- 
nesses of rosin-sized paper are placed between the wearing and 
the supporting wood. Where the timbers are on edge a layer 
(| in.) of tar cement (tar and sand) may be placed between 
the two sections. 



328 



GRAPHICS AND STRUCTURAL DESIGN 



Steel Floors. — The forms shown in Figs. 339 and 340 run to 
considerable weight and are suited to heavy loads. The fill 
usually consis s of a cinder concrete which need not be richer than 
1 part cement. 2 parts sand and 6 parts cinder. This is carried 

A. 




*p — r 

Fig. 340. 



a few inches above the top of the trough section. Sleepers are 
embedded in this fill and the wearing boards nailed to them. 

There are other trough sections made of much lighter metal, 
gauges No. 16 to No. 24. These sections are shown in Figs. 341 
to 343. Thesv. are made in both black and galvanized steel. 
The exposed side requires painting. 




Fig. 341. 



Fig. 342. 



Fig. 343. 



Brick Arch. — Another heavy floor construction is shown in 
Fig. 344. It consists of a common brick arch having a rise of 
preferably not less than one-eighth of the span. 

The tie rods should be placed at frequent intervals along the 
beams, especially along the outside beams, to prevent too great 
stress in the beams from lateral pressure. The horizontal thrust 
or pressure in pounds per lineal foot of arch is given in the Pen- 
coyd handbook as 

R 



SHOP FLOORS 



329 



W = load in pounds per square foot on the floor. 
S = span of arch in feet. 
R = rise of arch in inches. 



- " 

- '- : ■ - .-.:■■ ■- 




Fig. 344. 

The fill above the arch may be a lean cinder concrete, say, 
1-2-6. 

The wearing face may be any of those previously mentioned 
but is commonly maple. 




Fig. 345. 

The arch instead of being made of common brick may be any 
of the several types of hollow clay tile (Fig. 345), or may be a 
concrete arch placed on corrugated steel, expanded metal or 



gjH ^m^i 1 HHH 




Fig. 346. 



wire cloth; see Fig. 346. These floors are made suited to lighter 
floor loads than the first-described arch. 

The principal purpose of the corrugated steel, expanded metal 
or wire cloth is to carry the concrete during setting. The 



330 GRAPHICS AND STRUCTURAL DESIGN 

strength of these floors is due largely to the concrete. The 
thrust may be calculated by the formula given for brick arches. 

There are several forms of hollow-tile arches. The tile are 
made to give a flat arch when in place. The fill above the tile 
is a lean cinder concrete, Fig. 345. 

Reinforced-concrete Floors (Fig. 347). — These floors may be 
concrete slabs placed across steel beams as shown in the figure, 
or across reinforced-concrete beams, in which case the slab forms 
the flange of the T beam, the rectangular beam then becoming 
its stem. The design of these reinforced-concrete slabs and 
T beams has been fully considered in Chapter XIV. 



Fig. 347. 

The reinforcement may be either plain steel rods, twisted or 
deformed rods or any of the woven wires or expanded metals 
intended for such reinforcement. When rods are used some 
reinforcing steel should be placed at right angles to the main 
reinforcing rods. These slabs may have any of the previously 
described wearing surfaces. 

Concrete Wearing Surfaces. — This surface should be made to 
be as nearly waterproof as possible. The stone should not ex- 
ceed J in. ; it should be hard, either trap, granite, hard blue lime- 
stone or similar rock; no soft rocks are suitable. This concrete 
is frequently 1 part cement, 1 part sand, and 1 part stone; this 
should be laid upon the slab before the cement of the slab has 
fully set. Where the slab has fully set its face should be 
thoroughly cleaned and then covered with a neat cement slush 
upon which the wearing surface is then laid. 

Floors may be made practically waterproof by the thorough 
trowelling of the surface. It may also be waterproofed by coating 
it with pure linseed oil thinned with turpentine or naphtha. 



SHOP FLOORS 331 

Some concrete floors are painted but this, being merely a surface 
effect, soon wears off. 

Iron Floors. — Foundry floors, steel-mill charging floors and 
rolling-mill floors are frequently either steel or iron. These may 
be cast-iron plates, steel plates and in the case of steel-mill 
charging floors rolled-steel channels placed side by side upon 
their flange edges. 

Where a smooth surface is not one of the requirements of the 
floor, corrugated plates give a better footing. Several types of 
corrugated plates and of plates with holes in them are made by 
the various manufacturers of structural steel. 

Steps. — Concrete steps, when made of hard stone, wear well 
and give satisfaction. Corrugated-iron steps may be used and 
furnished with one of the patented edges to prevent slipping. 



CHAPTER XX 

WALLS AND ROOFS 

The ordinary forms of walls include: (i) Wooden walls or 
sides. (2) Corrugated-steel sides. (3) Brick, or brick with a 
concrete face. (4) Stone. (5) Solid-concrete walls in steel 
framing. (6) Hollow-concrete blocks. (7) Reinforced concrete 
with expanded metal or other reinforcement. 

1. Wooden walls or sides are serviceable where the nature of 
the work carried on in the building requires considerable addi- 
tional ventilation during part of the year, as the wooden sides 
may be readily removed. The wooden sides may be attached 
to either timber, metal or stone construction. 

2. Corrugated-steel Sides. — These are more substantial 
than the wooden ones. The common sizes of corrugated-steel 
sheets are given under roof coverings and the comments made 
there apply in a general way to the sides. The metal for the 
sides is ordinarily a little lighter than that on the roof, being 
commonly No. 20 or No. 22 gauge. 

The sheets may be secured to girts by clips or metal straps or 
may be nailed to studding. Where the building must be kept 
warm a lining of still lighter corrugated steel may be placed 
inside the first sheets, an air space being left between the two. 

3. Brick. — The required thickness of brick walls for any 
community is given by its "Building Code." An average state- 
ment of such requirements for a ten-story building would give 
about the following wall thicknesses. 

These sizes are approximate, being intended to be multiples 
of brick dimensions. They ordinarily apply to buildings not 
exceeding 120 ft. in length and 25 ft. in width. Where the walls 
exceed 120 ft. between lateral supports or the distance between 

332 



WALLS AND ROOFS 



333 



side walls is more than 25 ft. the thicknesses are commonly 
increased 4 ins. for each additional 120 ft. or part thereof in the 
length or each additional 25 ft. or part thereof in the width of 
the building. 



Numbers of stories. 


Estimated height 
in feet to roof. 


Thickness of wall 
in inches. 


10 

9 and 8 
8, 7 and 6 
5, 4 and 3 

2 and 1 


14 

40 

80 

120 

I50 


12-16 
16-20 
20-24 
24-28 
28-32 



Brick foundation walls or footings usually exceed the walls 
immediately upon them by at least 4 ins. The pressure per 
square inch upon the brick should not exceed 125 to 150 lbs. 

Where heavy concentrated loads, such as trusses or crane loads, 
are transferred through walls it is frequently more economical 
to buttress the wall at these points. In this case the design 
should be treated similarly to a pillar and the resultant pressure 
should fall within the middle third of the buttress. The walls 
between the buttresses can then be made lighter. 

Concentrated loads may also be carried by steel columns, 
either built in the wall or placed clear of the wall and inside the 
building. 

Where the walls are supported by columns, they need not 
exceed 12 ins. in thickness, while if the columns are not in the 
wall it should be made thicker. 

Where walls are cut into considerably for doors and windows 
their thicknesses should be increased. 

4. Stone walls are usually made thicker than brick walls would 
be made for the same situation. Few factory buildings are now 
made of stone unless very favorably located to suitable quarries. 

5. Solid-concrete Walls. — These are frequently light walls 
placed between the columns and steel framing of a building. 
The objections to a solid-concrete wall are the probability of 



334 



GRAPHICS AND STRUCTURAL DESIGN 



moisture passing through the concrete or the wail sweating on 
the inside, also the liability of such walls to crack when not 
reinforced to prevent it. Concrete walls of this type are usually 
made thinner than brick walls would be made for the same 
location. 

6. Hollow-concrete Blocks. — These are made in numerous 
forms. Being hollow they are poorer conductors of heat than 
solid walls and are therefore not liable to have moisture con- 
densed on the inside. Where desired, plaster can be placed 
directly on the blocks without furring and lathing. 

7. Reinforced-concrete Walls. — These may be made by 
running across the girts small f -in. channels, having flanges about 




Fig. 348. 



f in. wide, and spaced from 12 to 16 ins. center to center. 
Metal lathing is fastened to these channels, and upon this is 
placed a coating of concrete on both sides until the wall is made 
about 2 J ins. thick; see Fig. 348. 

When, on account of heating or dampness, a hollow wall is 
desired, an inside wall similar to the outside one but plastered 




Fig. 349. 



only on one side may be added, as in Fig. 349; an air space is 
left between the two wall sections. 

Through the courtesy of the American Bridge Company the 
following illustrations, Figs. 350 to 352, show the corrugated-steel 
details recommended by them, and Figs. 353 to 359 give their 
details of various types of window frames and sashes. 



-Roof Steel 



^-Purlin 



"^ 



lui Closing rivet 

V 

GABLE FINISH FOR STEEL END 




GABLE FINISH 

WITH PARAPET 

WALL 





Roof steel turned up 
behind vent end Bleel 

FINISH OF VENT ENDS 

Finish angle 



J* 

SIDE LAP FOR ROOF COR. ROOF 



TABLE OF CLINCH RIVETS 



Purlin leg 1 2 " 


■" 


_361 


7" 


Length 4" 


5" 


No.per pound] 48. 


U, 


33. 


27. 



Clinch rivets spaced every 6 
Rivet always to go through 
top of corrugations ~^^ j^' 



24 Gage 
Ridge roll 

is and bolta „ 
spaced every 6 



VALLEY GUTTER «>*' 

20 Gal. steel unless "f* -£»? 
otherwise specified 




Fig. 350. 

Corrugated steel for roofing is rolled from a sheet 30 inches wide in the flat, 27 £ inches 
when rolled one edge up and one edge down. Laid with x\ inches of corrugated lap will 
cover 24 inches of roof. 

When ordering state distinctly that the sheeting is for roofing, that it is to be 27 £ inches 
wide after corrugating, and that the corrugations are to be f inch deep. State whether the 
sheeting is to be plain or corrugated, and also if it is to be painted. Specify the gauge. 

Wherever possible order sheets in even feet lengths and to span two purlin spaces. 
Allow 6 inches for end laps in roofs of 6 inches pitch and 8 inches for end laps in roofs of 
4 inches pitch. In roofs of less than 4 inches pitch allow 8 inches for end laps and lay with 
Slater's cement. 

Sizes of Gutters and Conductors. 



Span of roof, feet. 



Up to 50 . 
50 to 70. 
70 to 100 




Conductor, inches. 



4 every 40 feet 

5 every 40 feet 
5 every 40 feet 



These are made of No. 24 galvanized steel unless otherwise specified, 
should slope at least 1 inch in 15 feet. 



Hanging gutters 



(335) 



53^ 



GRAPHICS AND STRUCTURAL DESIGN 



Cosing Rivet* 



Closing Rivet 



Clips and Bolts 

• PI 



Closing JUvets 

N3IDE CORNER 
CAPPING 




OUTSIDE CORNER 
CAPPING 



24 Net ^, 



Painted Cor. Steel 


Galv.Cor.Stee] 


Gage 


Weight 


Gage 


Weight 


16 


275 


16 


291 


18 


220 


18 


2o0 


20 


105 


20 


162 


22 


138 


22 


154 


24 


111 


24 


127 


20 


84 


26 


99 


27 


77 


27 


92 


28 


69 


28 


80 




Approximate Shipping Weight perSq, SIDE LAP FOR COR. STEEL SIDING 

Corrugated Sheets TJ.S.Standard 



Track and Hangers 
Head Jamb 



Wood Door Frame 




Fig. 351. 



Corrugated steel for siding is rolled from steel sheets 28 inches wide in the flat, and is 
26 inches wide when rolled with both edges down. When laid with a side lap of one corru- 
gation it will cover 24 inches of side. When ordering state distinctly that the sheeting is 
for siding, that it is to be 26 inches wide after corrugating, and that the corrugations are 
to be § inch deep. State whether the sheeting is to be galvanized or black, and if it is to 
be painted; also give gauge. Sheets should be ordered in even feet lengths and to span two 
purlin spaces, wherever possible. Allow 4 inches for end laps. If side laps are to be 
riveted space closing rivets about 12 inches apart. Provide roller guides and door stops 
to hold doors securely in place when open or shut. 



WALLS AND ROOFS 




( BERLIN 
LOUVRES 



Fig. 352. 



Shutters are made from 6 feet to 10 feet long, all standard width. Two hinges are used 
on shutters up to 8 feet long, while larger sizes are given three hinges. 

Louvres of the Berlin type are made of No. 24 gauge steel. The maximum length is 
4 feet 1 \ inches. The end lap is from I inch to f inch. 

Steel for these louvres should be ordered n inches wide. The uprights have %6-inch 
diameter holes for J-inch oval screw-head bolts, f inch long. 

Schiffler Louvres. The maximum lengths of these are 7 feet o inches, without lap. 
The steel should be ordered 11 inches wide. The holes are %.6 inch in diameter for i-inch 
diameter bolts with oval screw heads and 1 inch long. 



338 



GRAPHICS AND STRUCTURAL DESIGN 




-\ round -Ki 



1178 

-Glass- -~ft* — Glass- 

Wf=GlaBS,JIuntins andj:^- >-\ 

PLAN 

' Fig. 353- 



The design shown in Fig. 353 is for fixed sash for monitors; for swing monitor sash cut 
stops off as shown by dotted lines, and omit head stop on the inside. Make frames and 
sash of white pine, excepting spiking and blocking pieces, which should be spruce, hemlock 
or Norway pine, planed on all exposed sides. For swing sash order two trunnions for each 
sash, and call for lever operating device. 



WALLS AND ROOFS 



339 



Si" 




rlass-^j* — GJas3-*th~G.lass-->jt- 6 — Gdass- 

-W^Glass., Manttos,.and 4%-' 

PLAN 

Fig. 354- 



The design shown in Fig. 354 is for continuous fixed sash in corrugated steel sides, 
sash and sill of white pine, planed on all exposed sides. 



Make 



34Q 



GRAPHICS AND STRUCTURAL DESIGN 



Cor. Steel ,, „, 

Lag Screws&'Sc 1^" 
Flashing Spaced 2'to3'Ctrs. 
° /Purlin \ 





\-% Round 



-W = Glass, Muntins and 4J^— 
PLAN 

Fig. 355- 



The design shown in Fig. 355 is for sliding sash in corrugated steel sides. The stop 
used for roller track is to be hard wood, other wood the same as recommended for Fig. 353. 



WALLS AND ROOFS 



341 



xlSfLag Screw; 

lM"x3"x6"Block 

~T. 




Sill 
1%'la.g Screws 



Trim sheet under 
fluhdo.wJnJield 



ELEVATION 




l&Ml^-Glass- 



! 2 jk. . — ~~> -nK- -Glass- -*rf* — Glass- ■ > r r» | " l ' « - 1 j^ 
J*- W^Glass.Muntms.andlJjjWo'Max. 2 -&j | 
PLAN 

Fig. 356. 



The design shown in Fig. 356 is for a window frame with counterbalanced sash in corru- 
gated steel sides. Neither dimension of sash exceeds 4 feet o inches. The wood is recom- 
mended the same as for Fig. 353. 



342 



GRAPHICS AND STRUCTURAL DESIGN 



c 









Flashing 



Purlin 

— ■— g- 



Dimeu; 
by 



Top Rail 



fleeting Rai 



on of sash d 
ber and size 



iermined. 
>f lights" 



i^Cor.S 



Trim sheet under 
window in field 



ELEVATION 



4 



Z2 



P 



Purlin 



2x7 




• J&i 11%' \\% ZH"i 

^. • rs T — Glass-- -*fj«- - -Glass— -»ft< — Glass- 

f. — W =<Jlass,Muntins,and 4^£=>i'l Min. 

PLAN 

Fig. 357. 



H's. l%Lstg Screws 




tw 2x8Sill 
jfx^JiXag Screws 



SECTIO^ 



The design shown in Fig. 357 is for a window frame with counterbalanced sash in corru-^ 
gated steel sides. In this design one dimension is not less than 4 feet 1 inch. The wood* 
used is similar to that specified for Fig. 353. 



WALLS AND ROOFS 



343 



n m n 



.Cor.Steel 



-Flashing 



Purlin 



I 



q 



w 



Top Rail 



leeting Rai 



Dimens on of sash determined 
by num ber and size t f lights 



tfYim sheet under 
"window in field 



LCor.Steel 

ELEVATION 



H 



en 



k m 



m , ltf.x ±&* 
^ x % parting strip N \ Js'x^ 



%xl% Lag Screw 
2x-3"Block 




III Yk x 1 ^" Lag Screws 
IF 2x8''Sill 



%"x4Ji* 




^ ^ ■1 - Glass— ^J- 
1J§ r^-V^-Qlass, Muntins, and 4&=i'0 Max.- 



II 96 z X\tt-0--a I 

Glass-*))* 1 — &lass-*(<^-4>^t4-^ 

s, and 434=4' 0* Max.— H 

PLAN 

Fig. 358. 



344 



GRAPHICS AND STRUCTURAL DESIGN 



r 



r 



.Cor.Steel 

Flashing 



f-r lasning 

EJ Purlit 



w 



Top Rail 



Muntin 



fleeting Rail] 



n of sash determined 



by number and 6ize 



f lights 



Bottom Rail 



^= 






Purlin 



Cor.Steel 

ELEVATION 



H" We, 





Drip 
1-X-round 



Hxl% Lag Screw 
2 x 8*Sill 



SECTION 



M x % parting 6trip \ /i^' x £%' 



51 1 /» A *7a 



l-J^ round' 



!rr*K-*r 



Glass — Hp< — Glaes- 



2Kk if-s-ri 






1J£- i t*-W= Glass, Muntins, and 4}£=i 'film. -*j 

PLAN 

Fig. 359 



**V 



The designs shown in Figs. 358 and 359 are for window frames with double hung weighted 
sash in corrugated steel siding. In Fig. 358 neither dimension of the sash exceeds 4 feet 
o inches, while in Fig. 359 one dimension is not less than 4 feet 1 inch. The sills and cas- 
ings should be made of white pine; the jambs and parting strips should be made of hard 
pine; while spiking pieces and blocks should be made of hemlock or Norway pine, planed 
on all exposed sides. 



WALLS AND ROOFS 
CRANE CLEARANCES AND WEIGHTS 



34' 



I Not less than 3 " 




Capac- 




















ity, tons 
(2000 
lbs.). 


Span. 


A. 


B 




Wheel 
base. 


Weight of bridge, 
2 girders. 


Weight of 
trolley. 


Rail. 




Ft. 


Inches. 


Ft. 


Ins. 


Ft. 


Ins. 


Pounds. 


Pounds. 


Lbs. per 
yard. 


( 


40 


8 


4 


9 


8 


O 


8,600- 13,000 


4,500 


5 


60 


8 


4 


9 


8 


O 


16,500- 21,000 


to 


> 40 


( 


80 


8 


4 


9 


8 


O 


28,000- 33,000 


5,000 


) 


( 


40 


8 


5 





8 


6 


12,300- 16,000 


6,000 


) 


IO < 


60 


8 


5 





8 


6 


20,000- 24,000 


to 


> 40 


( 


80 


8 


5 





9 





32,000- 37,000 


8,000 


) 


( 


40 


8 


5 


6 


9 





14,000- 23,000 


9,000 


) 


15 


60 


8 


5 


6 


9 


6 


23,000- 33,000 


to 


50 


( 


80 


9 


5 


6 


10 





37,000- 48,000 


10,000 


) 


f 


40 
60 


9 
9 
9 
9 


6 









23,000- 28,000 
30,000- 37,000 
40,000- 52,000 
19,000- 34,000 


10,000 


\ 


20 < 


6 







to 


( 5 ° 


( 


80 


6 







12,000 


( 


40 


6 


6 


10 





12,000 


) 


25 


60 


9 


6 


6 


10 


6 


29,000- 45,000 


to 


} 60 


( 


80 


9 


6 


6 


11 





45,000- 61,000 


15,000 


) 


c 


40 


9 


7 





11 





34,000- 37,000 


14,000 


) 


30 < 


60 


9 


7 





11 


6 


44,000- 49, coo 


to 


} 60 


f 


80 


9 


7 





12 





58,000- 66,000 


17,000 


) 


( 


40 


9 


8 





12 


6 


43,000- 49,000 


16,000 


; 


40 < 


60 


9 


8 





13 





60,000- 63,000 


to 


> 80 


( 


80 


9 


8 





13 


3 


70,000- 82,000 


20,000 


5 


c 


40 


9 


8 


6 


12 


6 


48,000- 57,000 


24,000 


; 


50] 


60 


9 


8 


6 


13 





66,000- 73,000 


to 


} 100 


( 


80 


9 


8 


6 


13 


6 


85,000- 95,000 


30,000 


) 


( 


40 


10 


9 





15 





78,000 






60^ 


60 


10 


9 





IS 


3 


70,000- 95,000 


32,000 


^ 


f 


80 


10 


9 





15 


6 


100,000-126,000 




100 


c 


40 


10 


9 


6 


15 


6 


101,000 




75 


60 


10 


9 


6 


15 


6 


80,000-120,000 


40,000 


h to 


( 


80 


10 


9 


6 


15 


6 


120,000-144,000 






40 


12 


10 





is 


6 


134,000 




J 150 


100 < 


60 


12 


10 





15 


6 


94,000-161,000 


56,000 


I 


80 


12 


10 





is 


6 


125,000-187,000 







346 GRAPHICS AND STRUCTURAL DESIGN 

The maximum wheel loads are approximately 25 per cent of 
the bridge weight plus 50 per cent of the combined weights of the 
trolley and crane capacity. 

The dimensions of windows depend upon the light required 
and the size of the panes used. In shop construction the greatest 
possible amount of light is wanted but without sun. On this 
account a northern light is preferred. In one-story buildings this 
leads to saw-tooth or similar construction. 

The usual sizes of common glass can be obtained in any hand- 
book on Building Construction. The regular sizes of glass range 
from 6 X 8 ins. to 44 X 72 ins.; the dimensions up to 16 ins. 
varying by inches, while those above 16 ins. vary by 2 ins. and 
are the even inches only. 

Single-thick glass is about T X g in. thick. 

Double-thick glass is about ^ in. thick. 

Wired glass has the advantage of not falling out when cracked, 
thus affording increased fire protection. Translucent fabric is 
a substitute for glass that has some advantages. It cuts out a 
little light but when a sufficient amount is used it affords perfect 
lighting. It is impervious, elastic and burns with difficulty. 

The subject of methods of lighting is an interesting one and 
should be given careful consideration in buildings of any impor- 
tance. According to Ketchum it is common to specify that 10 
per cent of the exterior surface of ordinary mill buildings and 25 
per cent of machine shops and similar buildings should be glazed, 
this as a rule being divided between windows and skylights. 

Roof Coverings 

Corrugated-steel Roofing. — This is one of the commonest 
kinds of roohng materials. It is made by rolling the plain sheets 
with corrugations. 

These sheets range from No. 16 to No. 28, U. S. Standard 
gauge (0.063 i n - to 0.016 in.). The corrugations range from 
T 3 g in. to 2.5 ins. A few companies roll sheets with corrugations 
of 5, 3 and 2 ins. 



WALLS AND ROOFS 



347 



The sizes most frequently used are gauge Nos. 20 and 24, with 
2^-in. corrugations, | in. deep. The lengths of sheets range from 
6 ft. to 12 ft. A common length is 8 ft. 

Roofs covered with corrugated steel should not have a slope 
of less than 3 ins. in 12 ins. According to Kidder's Handbook 
they should be supported on about the following spans: 



Gauge. 


Span in feet. 


Gauge. 


Span in feet. 


24 

22 and 20 


2-2.5 
2-3 


18 
16 


4-5 
5-6 



The lap at the end of the sheet should range from 3 to 6 ins. 
according to the slope. 

Slope 1 to 3 Lap 3 ins. 

Slope 1 to 4 Lap 4 ins. 

Slope 1 to 8 Lap 5 ins. 

The side laps vary from one to two corrugations; a lap of i| 
corrugations makes a very good joint; see Fig. 360. 



Fig. 360. 



Corrugated roofing is secured to the building at the purlins 
by nails when nailing strips are used, and by hoop-iron clips or 
bands when secured directly to the purlins; see Figs. 350 to 352. 
The riveting and nailing should be done at the top of the corru- 
gations. 

The crippling load in pounds per square foot is approxi- 
mately 

/ 100,000 X t X d \ 
\ U I 



W = 



t = thickness of the metal in inches. 
d = depth of corrugations in inches. 
L = length of span in feet. 



348 



GRAPHICS AND STRUCTURAL DESIGN 



The following safe loads in pounds per square foot are one- 
quarter of the calculated crippling loads for sheets with 2! -in. 
corrugations, | in. deep. 







Span in feet. 




Gauge 










No. 












3 


4 


5 


6 


20 


65 


36 


23 


16 


22 


53 


30 


19 


13 


24 


43 


24 


IS 


II 


26 


32 


18 


12 


8 



Purlins should be spaced for a roof load of not less than 30 lbs. 
per sq. ft. ; spacing for lighter loads results in injury to the cor- 
rugated-steel joints when the roof is walked upon. The sizes 
most frequently used are No. 20 for the roof and No. 22 for 
the sides. The corrugated steel should be ordered in lengths 
sufficient to cover two purlins. The laps should be painted 
before being riveted, Galvanized steel will take paint better 
after weathering a while. 

When the buildings are lined, if by f-in. corrugated steel is 
frequently used. When lined the corrugated steel is better 
nailed to nailing strips. 

The greatest objection to the corrugated-steel roofing is the 
fact that it sweats. In heated buildings or in rooms with moist 
air the moisture condenses on the under side of the roof and 
drops to the floor, possibly striking machinery or materials it 
may injure. A common method of overcoming this condensation 
is to cover the purlins with galvanized poultry netting, running 
the netting from the eaves on one side, over the purlins and down 
to the eaves on the other side. The netting is secured to the 
eaves and the several widths of netting are woven together along 
the edges. Upon this netting are then laid one or two layers of 
asbestos paper, and on this one or two layers of tar or other 
paper impervious to moisture; finally the corrugated steel is 
placed on these. The netting is 60 ins. wide and weighs 10 lbs. 



WALLS AND ROOFS 



349 



per square of ioo sq. ft. The asbestos paper is about T X g in. 
thick. The asbestos paper prevents the tar or other paper from 
taking fire, thus making an excellent fireproof roof. 



Gauge. 


Weight of corrugated steel, 
in pounds per 100 sq. ft. 


Black. 


Galvanized. 


20 
22 
24 
26 


165 

138 

III 

84 


182 

154 
127 

99 



Slate Roofing. — Slate makes a very satisfactory roof cover- 
ing. The best slates have a somewhat metallic appearance, do 

not absorb water and are strong. 
Generally the strongest slates are 
the best. The sizes range from 
6 X 12 ins. to 16 X 26 ins. and in 
thickness from § in. to J in. The 
most frequently used sizes run from 
6 X 12 ins. to 12 X 18 ins., and 




Fig. 361. 



are generally ^ in. thick. In slate 
roofs the slope should not be less than i in 4. The usual lap is 
a double lap of 3 ins.; that is, the upper 3 ins. of the first slate 
are covered by the third slate; see Fig. 361. 

As slate breaks easily under shock, the roof must be designed 
to deflect but slightly under loading. 

The slates are commonly laid on sheathing; this maybe either 
plain or tongued-and-grooved boards, their thickness varying 
with the purlin spacing and the load. The sheathing is covered 
with tar paper or with waterproof paper or felt. The slates may 
be laid on roofing laths; these are 2 to 3 ins. wide and 1 in. to 
1 J ins. thick; they are spaced on the rafters to suit the nails in 
the slates. At gutters, valleys and ridges the slates should be 
laid in cement and are sometimes entirely laid that way. 



350 GRAPHICS AND STRUCTURAL DESIGN 

The following are examples of slate roofing: 

Trusses 14 ft. 6 ins. center to center. No purlins. Sheathing 
3 -in. yellow-pine plank. The planks were 29 ft. long and joints 
were broken on alternate trusses. 

Another roof whose purlins were 5 ft. 8 ins. center to center 
had 2 -in. yellow-pine sheathing. 

Although slate is an expensive roofing material, it can reason- 
ably be expected, with ordinary repairs, to last from 25 to 30 
years. While not resisting great heat (it cracks and disinte- 
grates, and then exposes the sheathing to the fire), slate itself 
does not ignite, and thus makes a fairly good fireproof roof. 

Clay tiles make a thoroughly fireproof roof. They are, however, 
very heavy and expensive. Several makes of tiles are also molded 
in glass, thus permitting skylights to be very readily placed. 
Their artistic effect is good. 

Concrete Roofs. — Concrete, reinforced by expanded metal 
or steel rods, is also much used for roofs. The purlins may be 
spaced from 5 to 7 ft., or they may be omitted altogether, the 
concrete slabs being placed directly on the trusses or on concrete 
beams or rafters. When the slope is steep, slate may be nailed 
directly to the concrete; this should be done within a couple of 
weeks after placing the concrete; or if the slope is slight a stand- 
ard slag roof may be used. 

The following is an example of a reinf orced-concrete roof : The 
roof slab was 4! ins. thick; it was reinforced with |-in. square 
bars, spaced 5 ins. center to center; there were no longitudinal 
beams; the transverse girders were 14 ft. center to center. 
Wooden nailing strips were embedded in the concrete slabs; 
these secured a five-ply waterproofing course upon which a stand- 
ard slag roof was placed. 

Slag or Gravel Roofing. — This roof covering may be placed 
on either wooden sheathing or on a reinforced-concrete roof. 
Such a roof when on wooden sheathing is commonly made by 
covering the wood with dry felt paper and on this placing from 
three to five layers of tarred or asphaltic felt. The layers of 



WALLS AND ROOFS 



351 



paper are lapped the same as slate, exposing from 6 to 10 ins. of 
each sheet. 

Another method is to lay rosin-sized sheathing paper weighing 
not less than 6 lbs. per 100 sq. ft.; upon this lay two ply of 
tarred felt lapping 17 ins. and weighing not less than 14 lbs. per 
100 sq. ft. single thickness. These are then covered with a spread- 
ing of pitch, and finally three additional layers of felt are laid and 
similarly coated with pitch. On this layer of pitch the slag or 
gravel is spread; if slag, it should be crushed smaller than f in. 
but larger than \ in. Gravel should be screened and both slag 
and gravel should be clean and dry. When laid on cement 
roofing the rosin sheathing paper should be omitted but the 
cement should be coated with pitch. Although a slag or gravel 
roof is not fire proof, tests have shown its fire-resisting qualities 
to be good. It protects the wooden sheathing better than slate. 



Slope of Roofs 

The slope of a roof depends upon that required by the roofing 
material used. The following is taken from Kidder's Handbook. 



Roofing material. 


Slope of rafter, 
inches per foot. 


Slate 


8 

7 
6 
6 

3 

1 


Tiles (interlocking) 


Tin shingles (painted) 


Cedar shingles 


Corrugated steel 


Ready roofing: 





In the case of slate the permissible slope will vary with the 
weight and size of the slate. 

Largest sizes of slate Slope 5 ins. per ft. 

Medium sizes of slate Slope 6 ins. per ft. 

Smallest sizes of slate Slope 8 ins. per ft. 

Roofs having a slope of § in. to f in. per ft. constitute what are 
termed flat roofs, and are generally covered with tar and gravel, 
asphalt, ready roofing or tin with lock and soldered joints. 



352 GRAPHICS AND STRUCTURAL DESIGN 

Pitch roofs should be given a slope not exceeding 2 ins. per ft. 

Rafters or their equivalents should be spaced not over 5 ft. 
when ordinary sheathing of i-in. boards is used. 

The following are examples of roofs : 

Pennsylvania Steel Co. — Purlins 6 ft. center to center. 
Sheathing if-in. tongued and grooved hemlock. Purlins 4 X 
10-in. yellow pine. Slag roofing, trusses 20 ft. center to center. 

A Bridge Shop Roof. — Composition roofing. Purlins spaced 
about 7 ft. 6 ins. center to center. Sheathing 2 ins. thick. 

Slate Roof on Tar Paper. — Purlins spaced about 4 ft. center 
to center. Sheathing i| in. 

Slate roof on 2-in. sheathing. Purlins spaced about 8 ft. center 
to center. 

Slate roof on 2^-in. plank. Purlins about 8 ft. 6 ins. center to 
center. 

Five-ply Slag Roofing. — Sheathing i-J-in. spruce on purlins 
spaced about 5 ft. 6 ins. center to center. 

Asphalt Roofing. — Purlins spaced 5 ft. 6 ins. center to center. 
Sheathing if-in. tongued and grooved boards. 



CHAPTER XXI 



SPECIFICATIONS FOR STRUCTURAL STEEL 

WORK 

MATERIALS 

i. Process of Manufacture. — Steel may be made by either the open- 
hearth or the Bessemer process. 

Note. — For the more important work rolled steel is preferably made by the 
basic open-hearth process, as this process permits the elimination of the greater 
portion of the phosphorus and a small percentage of the sulphur contained in the 
pig iron and scrap from which the steel is made. Steel for castings is commonly 
made by the acid open-hearth process. 

2. Chemical and Physical Requirements. — 



Chemical and physical requirements. 


Structural 
steel. 


Rivet steel. 


Steel castings. 


Phosphorus, maximum 


0.04% 
0.05% 

Desired 
60,000 

1,500,000 


0.04% 
0.04% 

Desired 

50,000 

1,500,000 


0.05% 
0.05% 

Not less than 
65,000 


Sulphur, maximum 

Ultimate tensile strength in 
lbs. per sq. in. See No. 3 . . . . 
Elongation, min. per cent in 8 




ins., see Fig. 1. See No. 4. . . . 
Elongation, min. per cent in 2 
ins., see Fig. 2 


Ult. ten. str. 

22% 
Silky 

j 180 flat 
) See No. 5 


Ult. ten. str. 

" Silky ' 

180 flat ) 
See No. 6 \ 


18% 

Silky or fine 

granular 

oo°, d = 3 / 


Character of fracture 


Cold bends without fracture .... 



Note. — The effect of phosphorus is to make the steel cold short, brittle when 
cold; while the sulphur makes it hot short, brittle when hot. 

3. Allowable Variations. — Tensile tests of steel will be considered 
satisfactory if showing an ultimate strength within 5000 pounds of that 
desired. 

4. Modifications in Elongation. — For material less than T \ inch thick 
a deduction of 2\ per cent will be allowed in the elongation for each ^ inch 

For material more than f inch thick a de- 
353 



the material is under A- inch. 



354 GRAPHICS AND STRUCTURAL DESIGN 

duction of i per cent may be allowed for each $ inch the material exceeds 
| inch. For pins and rollers exceeding 3 inches in diameter the elongation 
in 8 inches may be 5 per cent below that given in No. 2. 

5. Bending Tests. — These tests may be made either by pressure or by 
blows. Plates, shapes and bars less than 1 inch thick shall bend as specified 
in No. 2. 

Full-sized material for eyebars and other steel 1 inch thick and over, 
tested as rolled, shall bend cold 180 degrees around a pin whose diameter is 
twice the thickness of the bar. It must show no fracture on the outside 
of the bend. 

Angles I inch and less in thickness shall open flat, angles \ inch and less 
in thickness shall bend shut, cold, under blows of a hammer, without signs 
of fracture. This test will be made only when required by the inspector. 

6. Nicked Bends. — Rivet steel, when nicked and bent around a bar the 
same size as the rivet rod, shall give a gradual break and a fine, silky, 
uniform fracture. 

7. In order that the ultimate strength of full-sized annealed eyebars may 
meet the requirements of paragraph 65 the ultimate strength in test speci- 
mens may be determined by the manufacturers; all other tests than those 
for ultimate strength shall conform to the requirements in paragraph 2. 

8. The yield point, as indicated by the drop of the beam, shall be recorded 
in the test reports. 

9. Chemical Analyses. — Chemical determinations of the percentages 
of carbon, phosphorus, sulphur and manganese shall be made by the manu- 
facturer from a test ingot taken at the time of the pouring of each melt of 
steel and a correct copy of such analysis shall be furnished to the engineer 
or his inspector. Check analyses shall be made from the finished material, 
if called for by the purchaser, in which case an excess of 25 per cent above 
the required limits will be allowed. 



f j About 



1 



Parallel section 



About _j— -ij^ U i ; * ■ 



— 



» I *" . I T M I I ■ ■ ; 



Fig. 1. 



10. Form of Specimens. — (a) Plates, Shapes and Bars. — Specimens 
for tensile and bending tests for plates, shapes and bars shall be made by 
cutting coupons from the finished product, which shall have both faces 
rolled and both edges milled to the form shown in Fig. 1; or with both 



MATERIALS 



355 



edges parallel; or they may be turned to a diameter of f inch for a length 
of at least 9 inches, with enlarged ends. 

(b) Rivets. — Rivet rods shall be tested as rolled. 

(c) Pins and Rollers. — Specimens shall be cut from the finished rolled 
or forged bar in such a manner that the center of the specimen shall be 
1 inch from the surface of the bar. The specimen for the tensile test 
shall be turned to the form shown by Fig. 2. The specimen for bending 
test shall be 1 inch by \ inch in section. 




Fig. 2. 



11. Steel Castings. — The number of tests will depend on the number 
and importance of the castings. Specimens shall be cut cold from coupons 
molded and cast on some portion of one or more castings from each melt 
or from the sink heads, if the heads are of sufficient size. 

The coupon or sink head, so used, shall be annealed with the casting before 
it is cut off. Test specimens shall be of the form prescribed for pins and 
rollers. 

12. Annealed and Unannealed Specimens. — Material which is to be 
used without annealing or further treatment shall be tested in the condition 
in which it comes from the rolls. When material is to be annealed or 
otherwise treated before use, the specimens for tensile tests representing 
such material shall be cut from properly annealed or similarly treated short 
lengths of the full section of the bar. 

13. Number of Tests. — At least one tensile and one bending test shall 
be made from each melt of steel as rolled. In event of the material rolled 
from one melt varying in thickness by f inch or more a test shall be made 
from the thickest and the thinnest material rolled. 

14. Finish. — Finished material shall be free from injurious seams, 
flaws, cracks, defective edges, or other defects, and have a smooth, uniform, 
workmanlike finish. Plates 36 inches and under in width shall have rolled 
edges. 

15. Stamping. — Every finished piece of steel shall have the melt number 
and the name of the manufacturer stamped or rolled upon it. Steel for 
pins and rollers shall be stamped on the end. Rivet and lattice steel and 



356 GRAPHICS AND STRUCTURAL DESIGN 

other small parts may be bundled with the above marks on an attached 
metal tag. 

1 6. Defective Material. — Material, which, subsequent to the above 
tests at the mills, and its acceptance there, develops weak spots, brittle- 
ness, cracks, or other imperfections, or is found to have injurious defects, 
will be rejected at the shop and shall be replaced by the manufacturer at 
his own cost. 

17. Allowable Variation in Weight. — A variation in cross section or 
weight in the finished members of more than 2§ per cent from that specified 
shall be sufficient cause for rejection. 

18. Cast Iron. — Except where chilled iron is specified, castings shall 
be made of tough gray iron, with sulphur not over 0.10 per cent. They 
shall be true to pattern, out of wind (perfectly straight or flat) and free from 
flaws and excessive shrinkage. If tests are demanded they shall be made 
on the " Arbitration Bar " of the American Society of Testing Materials, 
which is a round bar i| inches in diameter and 15 inches long. The trans- 
verse test shall be on a supported length of 12 inches with the load at the 
middle. The minimum breaking load so applied shall be at least 2900 
pounds, with a deflection of at least -^0 mcn before rupture. 



Workmanship 

19. General. — All parts forming a structure shall be built in accord- 
ance with approved drawings. The workmanship and finish shall be equal 
to the best practice in modern bridge works. 

20. Straightening Material. — Material shall be thoroughly straightened 
in the shop, by methods that will not injure it, before being laid off or 
worked in any way. 

21. Finish. — Shearing shall be neatly and accurately done, and all 
portions of the work exposed to view shall be neatly finished. 

22. Rivets. — The size of rivets called for in the plans shall be under- 
stood to mean the actual size of the cold rivets before heating. 

23. Rivet Holes. — When general reaming is not required, the diameter 
of the punch for material not over f inch thick shall be not more than 
^s inch, nor that of the die more than | inch, larger than the diameter of 
the rivet. The diameter of the die shall not exceed the diameter of the 
punch by more than one-fourth the thickness of the material punched. 
Material over f inch thick, except that for minor details, and all material 
where general reaming is required, shall be sub-punched and reamed as per 
paragraphs 49, 50 and 51, or drilled from the solid. Holes in the flanges of 
rolled beams and channels used in the floors of railroad bridges shall be 



WORKMANSHIP 357 

drilled from the solid. Those in the webs of same shall be so drilled or sub- 
punched and reamed. 

Note. — Mr. Schneider in his specifications replaces the f inch occurring in 
paragraph 23 by f inch. 

24. Punching. — Punching shall be accurately done. Slight inaccuracy 
in the matching of holes may be corrected with reamers. Drifting to 
enlarge unfair holes will not be allowed. Poor matching of holes will be 
cause for rejection at the option of the inspector. 

Note. — Drifting is driving a taper pin through holes that fail to match properly; 
this distorts the hole and injures the material. 

25. Assembling. — Rivetedmembers shall have all parts well pinned 
up and firmly drawn together with bolts before riveting is commenced. 
Contact surfaces shall be painted (see paragraph 52). 

26. Lattice Bars. — Lattice bars shall have neatly rounded ends, unless 
otherwise called for. 

27. Web Stiffeners. — Stiffeners shall fit neatly between the flanges of 
girders. Where tight fits are called for, the ends of the stiffeners shall be 
faced and brought to a true contact bearing with the flange angles. 

28. Splice Plates and Fillers. — Web splice plates, and fillers under 
stiffeners, shall be cut to fit within | inch of flange angles. 

29. Connection Angles. — Connection angles for floor girders shall be 
flush with each other and correct as to position and length of girder. In 
case milling is required after riveting, the removal of more than r V inch from 
their thickness shall be cause for rejection. 

30. Riveting. — Rivets shall be driven by pressure tools wherever 
possible. Pneumatic hammers shall be used in preference to hand driving. 

31. Rivets. — Rivets shall look neat and finished, with heads of approved 
shape, full and of equal size. They shall be central on shank and shall 
grip the assembled pieces firmly. Recupping and calking will not be 
allowed. Loose, burned or otherwise defective rivets shall be cut out and 
replaced. In cutting out rivets great care shall be taken not to injure the 
adjacent metal. If necessary they shall be drilled out. 

31 (a). Heating Rivets. — Rivets shall be heated to a light cherry red, 
in a gas or oil furnace. The furnace must be so constructed that it can 
be adjusted to the proper temperature. 

Note. — Paragraph 31 (a) is inserted by Mr. Schneider but is not generally in- 
cluded by others. 

32. Field Bolts. — Wherever bolts are used in place of rivets which 
transmit shear, the holes shall be reamed parallel and the bolts turned to 
a driving fit. A washer not less than \ inch thick shall be used under the 
nut. 



358 GRAPHICS AND STRUCTURAL DESIGN 

33. Members to be Straight. — The several pieces forming one built-up 
member shall be tight and fit closely together, and finished members shall 
be free from twists, bends or open joints. 

34. Finish of Joints. — Abutting joints shall be cut or dressed true and 
straight and fitted closely together, especially where open to view. 

In compression joints, depending on contact bearing, the surfaces shall 
be truly faced, so as to have even bearings after they are riveted up com- 
plete and perfectly aligned. 

35. Field Connections. — Holes for floor-girder connections shall be sub- 
punched and reamed with twist drills to a steel template 1 inch thick. 
Unless otherwise allowed, all other field connections shall be assembled in 
the shop and the unfair holes reamed; when so reamed the pieces shall be 
match-marked before being taken apart. 

36. Eyebars. — Eyebars shall be straight and true to size, and shall be 
free from twists, folds in the neck or head, or any other defect. 

Heads shall be made by upsetting, rolling or forging. Welding will not 
be allowed. The form of the heads will be determined by the dies in use 
at the works where the eyebars are made, if satisfactory to the engineer, 
but the manufacturer shall guarantee the bars to break in the body with 
a silky fracture, when tested to rupture. The thickness of the head and 
neck shall not vary more than ^ inch from that specified. 

37. Boring Eyebars. — Before boring each eyebar shall be properly 
annealed and carefully straightened. Pinholes shall be in the center line 
of the bars and in the center of the heads. Bars of the same length shall 
be bored so accurately that, when placed together, pins fc inch smaller 
in diameter than the pinholes can be passed through the holes at both ends 
of the bars at the same time. 

38. Pinholes. — Pinholes shall be bored true to gauges, smooth and 
straight, at right angles to the axis of the member and parallel to each 
other, unless otherwise called for. Wherever possible, the boring shall be 
done after the member is riveted up. 

39. Variation in Pinholes. — The distance center to center of pinholes 
shall be correct within ¥ V mcn > an d for pins up to 5 inches in diameter the 
diameter of the hole shall not exceed the diameter of the pin by more than 
Jq inch; for larger pins this difference shall not exceed ¥ V inch. 

40. Pins and Rollers. — Pins and rollers shall be accurately turned to 
gauges and shall be straight, smooth and entirely free from flaws. 

41. Pilot Nuts. — At least one pilot and driving nut shall be furnished 
for each size of pin for each structure, and field rivets 10 per cent in excess 
of the number of each size actually required. 

Note. — Pilot and driving nuts are nuts used to guide and protect truss pins 
during driving on erection. 



PAINTING 359 

42. Screw Threads. — Screw threads shall make tight fits in the nuts 
and shall be U. S. standard, except above the diameter of if inches, when 
they shall be made with six threads per inch. 

43. Annealing. — Except in minor details, steel which has been partially 
heated shall be properly annealed. 

44. Steel Castings. — All steel castings shall be annealed. 

45. Welds. — Welds in steel will not be allowed. 

46. Bed Plates. — Expansion bed plates shall be planed true and smooth. 
Cast wall plates shall be planed top and bottom; the cut of the planing 
tool shall correspond with the direction of expansion. 

47. Shipping Details. — Pins, nuts, bolts, rivets and other small details 
shall be boxed or crated. 

Additional Specifications when General Reaming and Planing 

are Required 

48. Planing Edges. — Sheared edges and ends shall be planed off at 
least \ inch. 

49. Reaming. — Punched holes shall be made with a punch T 3 g inch 
smaller in diameter than the nominal size of the rivets and shall be reamed 
to a finished diameter of not more than xV inch larger than the rivet. 

50. Reaming after Assembling. — Wherever practicable, reaming shall 
be done after the pieces forming one built member have been assembled and 
firmly bolted together. If necessary to take the pieces apart for shipping 
and handling, the respective pieces reamed together shall be so marked that 
they may be reassembled in the same position in the final setting up. No 
interchange of reamed parts will be allowed. 

51. Removing Burrs. — The burrs on all reamed holes shall be removed 
by a tool countersinking about xV inch. 

Painting 
51 (a). Shop Painting. — Steel work, before leaving the shop, shall be 
thoroughly cleaned and given one good coating of pure linseed oil, or such 
paint as may be called for, well worked into all joints and open spaces. 

52. In riveted work, the surfaces coming in contact shall each be painted 
before being riveted together. 

53. Pieces and parts which are not accessible for painting after erection, 
including tops of stringers, eyebar heads, ends of posts and chords, etc., 
shall be given two coats of paint before leaving the shop. 

54. Steel work to be entirely embedded in concrete shall not be painted. 

55. Painting shall be done only when the surface of the metal is per- 
fectly dry. It shall not be done in wet or freezing weather, unless protected 
under cover. 



360 GRAPHICS AND STRUCTURAL DESIGN 

56. Machine-finished Surfaces. — These shall be coated with white 
lead and tallow before shipment or before being put out into the open air. 

57. Field Painting. — After the structure is erected, the metal work shall 
be painted thoroughly and evenly with an additional coat of paint, mixed 
with pure Unseed oil, of such quality and color as may be selected. 

Succeeding coats of paint shall vary somewhat in color, in order that 
there may be no confusion as to the surfaces that have been painted. 

Inspection and Testing 

58. Facilities for Inspection. — The manufacturer shall furnish all 
facilities for inspecting and testing the weight, quality of material and 
workmanship at the shop where the material is manufactured. He shall, 
if required, furnish a suitable testing machine for testing full-sized members. 

The manufacturer shall prepare all test pieces for the machine, free 
of cost. 

59. Access to Shop. — When an inspector is furnished by the purchaser, 
he shall have full access, at all times, to all parts of the shop where material 
under his inspection is being manufactured. 

60. The purchaser shall be furnished complete shop plans, and must be 
notified well in advance of the start of work in the shop, in order that he 
may have an inspector on hand to inspect material and workmanship. 

61. Shipping Invoices. — Complete copies of shipping invoices shall be 
furnished to the purchaser with each shipment. 

62. Mill Orders. — The purchaser shall be furnished with complete 
copies of mill orders, and no material shall be rolled and no work done before 
he has been notified as to where the orders have been placed, so that he may 
arrange for the inspection. 

63. Inspector's Mark. — The inspector shall stamp with a private mark 
each piece accepted. Any piece not so marked may be rejected at any 
time, at any stage of the work. If the inspector, through an oversight or 
otherwise, has accepted material or work which is defective or contrary to 
specifications, this material, no matter in what stage of completion, may be 
rejected by the purchaser. 

Full-sized Tests 

64. Full-sized Tests. — Full-sized parts of the structure may be tested 
at the option of the purchaser. Such tests on eyebars and similar members, 
to prove the workmanship, shall be made at the manufacturer's expense, 
such members shall be paid for by the purchaser, at contract price, if the 
tests are satisfactory. If the tests are not satisfactory, the members repre- 
sented by them will be rejected. The expense of testing members, to prove 
their design, shall be paid for by the purchaser. 



STEEL MILL BUILDINGS 36 1 

65. Eyebar Tests. — In eyebar tests, the minimum ultimate strength 
shall be 55,000 pounds per square inch. The elongation in 10 feet, in- 
cluding fracture, shall be not less than 15 per cent. Bars shall break in 
the body and the fracture shall be silky or fine granular. The elastic limit 
as indicated by the drop of the mercury shall be recorded. Should a bar 
break in the head and develop the specified elongation, ultimate strength 
and character of fracture, it shall not be cause for rejection, provided not 
more than one-third of the total number of bars break in the head. 

SPECIFICATIONS FOR STEEL MILL BUILDINGS 

66. Dimensions. — By height shall be understood the distance from the 
under side of the lower chord of the roof truss to the tops of the foundations. 

The width and length of the building shall be measured to the outsides 
of the framing or sheathing. 

67. Spans. — The spans of trusses, beams and girders for the purpose of 
calculations shall be assumed as the distance center to center of bearings 
for supported spans, while they will be considered as the distance back to 
back of framing angles when the trusses, beams or girders are framed into 
columns. 

Loads 

68. Dead Loads. — The dead load is the weight of all permanent con- 
struction and fixtures. The calculated weights shall be based upon those 
hereafter given. 

(a) Weight of Trusses. — The weight of a truss may be estimated as 

P-L 



W 



^—\ 



3 
where 

W = weight of truss per square foot of building area, 
L = span of truss, in feet, 

D = distance center to center of trusses, in feet, 
P = load per square foot on the truss, 
(b) Weight of Purlins. — The weight of purlins per square foot of 
horizontal projection of the roof may be taken as 

45 4 

where 

Wi = weight of purlins per square foot of building, 
D = distance center to center of the trusses, 
Pi = load per square foot on purlins. 



362 GRAPHICS AND STRUCTURAL DESIGN 

(c) Weight of Roof Coverings. — The weights of roof coverings not 
including sheathing per square foot of actual roof surface will average 

Tin 1 lb. Corrugated steel, No. 18 ... . 3.0 lbs. 

Slate, xVmch 6.6 lbs. Corrugated steel, No. 20 ... . 2.3 lbs. 

Slate, £-inch 4.4 lbs. Felt and gravel 7-9 lbs. 

Terra-cotta, 1 inch thick. . 6.0 lbs. 

(d) Weight of Sheathing. — 

Wooden sheathing, per foot, board measure 3.5 lbs. 

Concrete, 1 inch thick, per square foot 10-12 lbs. 

Note. — A foot board measure is a piece 1 2 inches square and 1 inch thick. 

Live Loads 

69. (a) Live Loads. — These shall not be less than the following uniform 
loads : 

Warehouses 120 pounds per square foot. 

Foundry charging floors 300 pounds per square foot. 

Power houses, uncovered floors . . . 200 pounds per square foot. 

In all cases special concentrations, such as engines, turbines, boilers, chim- 
neys, etc., should be considered from actual weights. 

(b) Crane Loads. — Where available the actual weights and dimensions of 
traveling cranes should be used; otherwise the table of weights given on page 
364 and taken from the Specifications of Mr. C. C. Schneider may be used. 

(c) Distribution of Wheel Loads. — Wheel loads transferred to beams or 
girders through rails may be considered as distributed a distance equal to 
the depth of the girder or beam but not exceeding 30 inches. (Ostrup.) 

(d) Lateral Loading. — Besides the vertical loading, beams and girders 
carrying traveling cranes shall be designed so that their upper flanges shall 
in addition resist a lateral bending due to one-twentieth of the crane's 
capacity acting at each bridge-truck wheel. 

(e) Impact. — An addition of 25 per cent of the live-load bending mo- 
ments and shears shall be made to all beams, girders and columns carrying 
traveling cranes. 

70. Flat Roofs. — Flat roofs liable to be loaded with people should be 
designed as floors. This would subject them to a possible additional load 
of 40 pounds per square foot. 

71. Wind Loads. — The normal wind pressure on a roof shall be based 
upon a horizontal wind pressure of 20 pounds per square foot and its mag- 
nitude shall be estimated by Duchemin's formula, 

1 + sm 2 A 



STEEL MILL BUILDINGS 363 

where W n is the normal component of the wind pressure in pounds per 
square foot. 

Wh is the horizontal pressure in pounds per square foot on a vertical 

surface. 
A is the slope of the roof with the horizontal in degrees. 
For Wh = 2 °> W n has the following values. 

A w n A w n 

5 3-5 33-7 (slope 2 to 3) 17.0 

10 6.7 40 18 

20 12.3 45 18 

21.8 (slope 2 to 5) 13. 1 50 19 

26.6 (slope 1 to 2) 14. 9 60 19 

30 16.0 90 ... 20 

Note. — The available data relating to wind pressures on structures is exceed- 
ingly unsatisfactory. This relates both to the horizontal pressure corresponding to 
any wind velocity and the effect on the large surfaces exposed by ordinary build- 
ings. Of the several formulas proposed for estimating the normal wind pressure 
upon inclined surfaces Duchemin's gives values somewhat higher than Hutton's 
or the straight-line formula and, as it agrees with recent small surface tests, is pre- 
ferred by many. 

Hutton's formula, which was formerly generally accepted, is 

W n = W h sin A*- 8 * 2 cos A -I. 

The straight-line formula, which gives results closely approximating those 
given by Hutton's formula, is 

WhXA ■ 



W n = 



45 



In this last formula A is expressed in degrees and the formula only applies 
to angles of slope not exceeding 45 degrees. The specifications of the 
American Bridge Co. require that "the wind pressure shall be assumed 
acting horizontally, at 20 pounds per square foot, on the vertical projec- 
tions of roof surfaces." Still other specifications assume a horizontal wind 
pressure of 30 pounds per square foot reduced to normal pressure by Du- 
chemin's formula. 

Observations made by M. E. Bret on the top of the Eiffel Tower and 
recorded in "Le Genie Civil," March 12, iqio, led him to conclude that 
horizontal wind pressure seldom exceeded 20 pounds per square foot, and 
that the relation between pressure and velocity should read p = 0.0029 v 2 , 
where p = pressure per square foot in pounds and v = the velocity of the 
wind in miles per hour. 



3 6 4 



GRAPHICS AND STRUCTURAL DESIGN 



It is generally accepted that wind pressures per square foot are lower over 
large areas than over small ones, and that these pressures are lower near 
the ground than at higher altitudes. 

Although the experiments of Stanton, Smith, Boardman and others seem 
to indicate the existence of suction acting upon the leeward sides of buildings, 
walls and roofs, their tests were upon too small scales to warrant the im- 
mediate changing of usual practice. 



WEIGHTS AND DIMENSIONS OF TYPICAL TRAVELING 

CRANES 



Capacity. 


Span. 


Wheel base. 


Maximum 
wheel load. 


Side 
clearance. 


Vertical 
clearance. 


Weight of rail for 


Tons. 


Feet. 


Ft. Ins. 


Pounds. 


Inches. 


Ft. Ins. 


Runway 
girder. 


Beams. 


5 

5 


40 
60 


8 6 

9 


12,000 
13,000 


IO 
IO 


6 
6 


40 
40 


40 
40 


IO 
IO 


40 
60 


9 O 
9 6 


19,000 
21,000 


IO 
IO 


6 
6 


45 
45 


40 
40 


15 
15 


40 
60 


9 6 
10 


25,000 
29,000 


IO 
IO 


7 
7 


5o 

5o 


50 
50 


20 
20 


40 
60 


10 
10 6 


33,000 
36,000 


12 
12 


7 
7 


55 
55 


50 
50 


25 
25 


40 
60 


10 
10 6 


40,000 
44,000 


12 
12 


8 
8 


60 
60 


50 
50 


30 
30 


40 
60 


10 6 • 

11 


48,000 
52,000 


12 
12 


8 
8 


70 
70 


60 
60 


40 
40 


40 
60 


11 

12 


64,000 
70,000 


12 
12 


9 
9 


80 
80 


60 
60 


5o 
5o 


40 
60 


11 

12 


72,000 
80,000 


14 
14 


9 
9 


100 
100 


60 
60 



Note. — The rail weights are in pounds per yard. The side clearance is from 
the center of the rail, while the vertical clearance is from the top of the rail. 



72. Wind Pressure on the Sides and Ends of a Building. — The wind 
pressure on the sides and ends of a building shall be taken at 20 pounds per 
square foot. 

Note. — Other specifications assume 20 pounds per square foot on buildings 
not exceeding 30 feet to the eaves and this is increased up to 30 pounds per square 



STEEL MILL BUILDINGS 365 

foot for buildings measuring 60 feet or more to the eaves. Still other sped-- 
fications for positions not severely exposed substitute 15 pounds per square foot 
for the 20 pounds specified in the first sentence of paragraph 72, and where the 
distance from the ground to the eaves exceeds 25 feet the normal pressure on 
the roof is that corresponding to a horizontal pressure of 20 pounds per square 
foot. 

73. Wind Pressure on Framework. — The wind pressure shall be esti- 
mated for the wind acting horizontally in any direction upon the total 
exposed surface of the framework. Such pressure shall be assumed at 
30 pounds per square foot of exposed surface. (This applies during con- 
struction.) 

74. Snow Loads. — For latitudes of about 40 degrees the snow load 
may be assumed at 25 pounds per square foot of horizontal projection of 
the roof for flat roofs and those inclined up to angles of 20 degrees; above 
this slope the load may be decreased uniformly, so that at 45 degrees it 
would become zero. (The slope corresponding to 20 degrees is about one 
in three.) 

These loads should be increased for higher latitudes and reduced for 
lower latitudes. In tropical countries snow loads may be neglected. 

The amount and character of the precipitation in any locality may 
also affect the loading, which would naturally be lower in arid sec- 
tions. 

Note. — According to Mr. C. C. Schneider, in climates corresponding to that of 
New York, ordinary roofs up to spans of 80 feet may be designed for the following 
minimum equivalent loads per square foot of exposed roof surface. These loads 
then replace the dead loads, wind loads and snow loads given above. 

Lbs. 

Gravel or f On boards, slope 1 to 6, or less 50 

composition \ On boards, slope exceeding 1 to 6 45 

roofing [ On 3-inch flat tile or cinder concrete 60 

Corrugated sheets, on boards or purlins 40 

Slate, on boards or purlins 50 

Slate, on 3-inch flat tile or cinder concrete 65 

Tile on steel purlins 55 

Glass 45 

Where no snow is to be expected the above loads may be reduced 10 pounds, 
excepting that no roof should be designed for a load less than 40 pounds per square 
foot. 

74a. Trusses Forming Parts of Bents. — Truss members shall be de- 
signed for those combinations of dead load, snow load and wind load 



366 GRAPHICS AND STRUCTURAL DESIGN 

which shall require maximum sections. The three possible combinations 
shall be assumed, 

1. Dead load plus maximum snow load. 

2. Dead load plus wind load. 

3. Dead load plus minimum snow load plus wind load. 

The minimum snow load is generally assumed as one-half the maximum 
snow load. It is commonly considered that the full snow load will not act 
simultaneously with the maximum wind load. 

Note. — Where trusses are placed upon columns, forming bents, and the struc- 
ture is stiffened to resist the wind pressure by the introduction of knee-braces be- 
tween the trusses and the columns, the stresses, determined upon the basis of an 
equivalent uniform load, give no indication of the effect of the wind pressure upon 
the members of the trusses or upon the columns. When designed in this way 
careful consideration should be given to the effect of wind upon the structure, 
particularly regarding the possibility of compression in pieces subjected to tension 
for uniform loads. Such members should preferably be made of two angles having 
a suitable l/r value. 

75. Minimum Loads on Purlins and Roof Coverings. — Purlins and roof 
coverings shall not be designed for normal loads under 30 pounds per square 
foot. 

76. Loads on Foundations. — The areas of foundation piers shall be 
proportional to their respective dead loads; in no case, however, shall the 
combined live and dead loads on a pier exceed the permissible pressure on 
the soil. 

Note. — The desire is to obtain uniform settlement. In making the founda- 
tions proportional to the dead loads it is considered that as the dead loads act 
continuously the settlement will depend more on dead loads than on the possibly 
very intermittent action of the live loads. 



UNIT STRESSES AND PROPORTION OF PARTS 

Substructure 

77. Pressures on Soils. — The pressures on the soil at the base of the 
foundation shall not exceed the following in tons of 2000 pounds. 

Clay, soft 1 

Ordinary clay, or dry sand mixed with clay 2 

Dry sand and dry clay 3 

Hard clay and firm coarse sand 4 

Firm gravel and coarse sand 6 

Rock, according to condition 15-200 



STEEL MILL BUILDINGS 367 

78. Compressive Stresses in Masonry. — The following compressive 
stresses will be permitted in masonry structures: 

Lbs. per sq. in. 

Common brick, in Portland cement mortar. ........ 170 

Hard-burned brick, in Portland cement mortar 200 

Rubble masonry, in Portland cement mortar 150 

Sandstone masonry, first class 280 

Limestone masonry; first class 350 

Granite masonry, first class 420 

Portland cement concrete, 1-2-4 4°° 

Portland cement concrete, 1-2-5 3°° 

79. Wall-plate Pressure. — According to Mr. C. C. Schneider, the 
pressure of beams, girders, wall plates, columns, etc., on masonry shall not 
exceed the following: 

Lbs. per sq. in. 

On brick work in cement mortar 300 

On rubble masonry in cement mortar 250 

On Portland cement concrete (1-2-4) 600 

On first-class sandstone, dimension stone : 400 

On first-class limestone 500 

On first-class granite 600 

80. Bearing Power of Piles. — Piles shall not be spaced closer than 30 
inches center to center. The maximum load on any pile shall not exceed 
40,000 pounds, or 600 pounds per square inch of average cross section. 
When driven to rock or equivalent bearing through loose or wet soil, which 
gives them no lateral support, the limiting load shall be determined by 
reducing 600 pounds, the maximum allowable compression, by a suitable 
column formula. 

81. Walls shall be built in accordance with the local "building code" 
when that is available; otherwise they may be made the thicknesses given 
on page 333. 

82. Pillars. — (a) When concentrically loaded and having a height not 
exceeding twelve times their least dimension, pillars may be loaded until 
the fiber stresses reach the figures given in paragraph 78. 

(b) When eccentrically loaded the resultant pressure must pass within 
the kern of the section and the maximum pressure should not exceed that 
given in paragraph 78. 

Note. — For the explanation of kern of a section see page 247. 



3 68 



GRAPHICS AND STRUCTURAL DESIGN 



Unit Steesses in Steel Work 

83. Permissible Stresses. — The resulting stresses due to dead load, 
snow load, wind load and impact shall not exceed the following limiting 
values, excepting where permitted in accordance with paragraph 91. 

84. Tension, net section, rolled steel, 16,000 pounds per square inch. 

85. Direct Compression, rolled steel and steel castings, 16,000 pounds 
per square inch. 

86. Bending Stresses, on extreme fibers of rolled shapes, built sections, 
girders and steel castings, net section, 16,000 pounds per square inch. 

On extreme fibers of pins, 24,000 pounds per square inch. 

87. Shearing Stresses. — Lbs. per sq. in. 

On shop rivets and pins 1 2,000 

On bolts and field rivets 10,000 

On plate girders and beams, cross section, average 10,000 

88. Bearing Stresses. — Lbs. per sq. in. 

- On shop rivets and pins 24,000 

On field rivets and pins 20,000 

89. Pressure on Expansion Rollers. — The pressure per linear inch on 
expansion rollers shall not exceed 600 pounds per inch of roller diameter. 

90. Axial Compression in Columns. — The load per square inch of gross 

section for columns axially loaded shall neither exceed 16,000 — (C-) 

pounds, nor 14,000 pounds. Here I = the length of the member in inches; 
r = the corresponding radius of gyration of the section in inches; and C 
has the following values: 



Condition of ends. 


C. 


Both ends hinged or butting 


70 

35 

47 

140 


Both ends fixed .• 


One end fixed, the other hinged 


One end fixed, the other free 





Note. — The specifications issued Dec. 1, 1912, by the American Bridge Co. 
permit loads upon columns, concentrically loaded, as given by the formula 19,000 — 

( 100 X - J but not to exceed 13,000 pounds per square inch and applying to values 



of - up to 120. 
r 



STEEL MILL BUILDINGS 



3 6 9 



For values of - from 120 to 200 they give the following table: 
r 



I 


Allowable load, 


/ 


Allowable load, 


r 


lbs. per sq. in. 


r 


lbs. per sq. in. 


120 


7000 


160 


5000 


130 


6500 


170 


4500 


140 


6000 


180 


4000 


150 


5500 


190 


3500 






200 


3000 



91. Stresses in Bracing and Members with Combined Stresses. — In 
bracing and members subjected to combined stresses resulting from wind 
and other loadings the permissible working stresses previously given may 
be increased 25 per cent, but the section shall not be less than that other- 
wise required by the other loadings excluding the wind load. 



92. Eccentric Loading, 
for eccentric loading. 



In designing columns provision must be made 



93. Combined Stresses. — Members subjected to combined axial and 
bending stresses shall be designed so that the greatest fiber stress due to 
the combined stresses shall not exceed the fiber stress permitted in the 
member. 

94. Alternate Stresses. — Members subjected to a reversal of the stresses 
shall be proportioned for that stress giving the greater section. The con- 
nections shall be designed to carry the sum of the stresses. 

95. Net Sections. — In finding net sections, the rivet holes shall be as- 
sumed as I inch larger than the nominal diameter of the rivet. 

96. Limiting Lengths of Columns. — Columns assumed as having hinged 
or butting ends shall have unsupported lengths not exceeding 125 times their 
least radius of gyration for main members nor more than 150 times their 
least radius of gyration for wind bracing or secondary members. 

Note. — The recent specifications of the American Bridge Co. place these lengths 
at 120 and 200 times the least radjus of gyration. 

97. Limiting Lengths of Tension Pieces. — Riveted tension pieces shall 
have lengths not exceeding 200 times their radii of gyration about their 
horizontal axes, when used in horizontal or inclined positions. The 
length for this calculation shall be considered that of its horizontal pro- 
jection. 



370 GRAPHICS AND STRUCTURAL DESIGN 

98. Rolled Sections used as Beams. — Rolled sections used as beams 
shall be proportioned by their moments of inertia. 

99. Plate Girders. — Plate girders shall be designed upon the assump- 
tion that one-eighth of the web acts as flange area. 

Note. — For buildings and similar structures it is frequently specified that the 
bending stresses shall be resisted by the flanges and that the web shall resist shear 
only. 

Note. — The recent specifications of the American Bridge Co. permit the design 
of girders by their moments of inertia, thus making paragraph 99 apply to both 
girders and rolled sections. 

100. Web Thickness of Plate Girders. — The web thickness shall be not 
less than T ^o of the unsupported distance between flange angles. 

101. Web Stififeners. — Web stiff eners shall be placed on both sides of 
the web, with a close bearing against upper and lower flange angles, at the 
ends and inner edges of bearing plates, at all local and concentrated loads; 
where the thickness of the web plate is less than ^ of the unsupported 
distance between flange angles, stiffeners shall be placed at intervals along 
the girder about equal to the girder depth but not exceeding 5 feet. 

Note. — The recent specifications of the American Bridge Co. place this limiting 
distance at 6 feet. 

102. Compression Flange. — The area of the compression flange must 
at least equal the area of the tension flange at the same section of the girder. 
Where the unsupported distance along the girder exceeds 15 times the width 
of the flange the fiber stress in the compression flange must be that deter- 
mined by considering the flange as a column and buckling laterally. 

Note. — The practice on this point varies considerably among designers. Mr. 
C. C. Schneider limits the fiber stress in the compression flanges at all times to 
that given by the following formulas. When the flange section consists of angles 

and plates the fiber stress is limited by 16,000 — f 200 X 7 ) , and where the flange 

is a channel section by 16,000 — f 150 X t )• Here I is the distance along the 

flange between points of support and b is the width of the flange plate, both being 
in inches. 

The requirements as stated by the specifications of the American Bridge Co. are 
that the flange must be braced at intervals not exceeding 40 times the width of 
the flange plates, and that the limiting fiber stress for unsupported flange lengths 
exceeding 10 times the widths of the flanges shall be that given by 19,000 — 

( 300 X 7 ). The symbols have the same significance as those given above. 

Other specifications permit an unsupported flange length of 12, 16 and even 
20 times the width of the flange plates before requiring any reduction in the fiber 
stress. 



STEEL MILL BUILDINGS 37 1 

103. Depths of Rolled Beams. — The depths of rolled beams where used 
for floors shall be not less than J$ of the span; when used for purlins they 
may be as small as gV of the span. Floor beams when subjected to con- 
siderable vibration or shock should be made not less than y^ of their span. 

Note. — The purpose of this requirement is to limit the deflection. 

104. Permissible Stresses on Cast Iron. — 

Tension 2,500 lbs. per sq. in. 

Compression 12,000 lbs. per sq. in. 

Shear 1*500 lbs. per sq. in. 

105. Permissible Stresses on Timber. — The timber in the structure 
may be designed with the following fiber stresses given in pounds per square 
inch. 



White oak 

Long-leaf yellow pine . 
White pine and spruce 
Hemlock 



Bending 

stress. 



I200 

1500 

900 

700 





Columns under 






10 diameters. 


Shear 


Bearing 




along 


on end. 


Com- 


Bearing 


fibers. 




pres- 


across 






sion. 


fibers. 




1400 


IOOO 


500 


200 


1500 


IOOO 


350 


IOO 


900 


700 


200 


IOO 


700 


SOO 


200 


IOO 



Weight 

per 
cu. ft. 



SO 
38 
24 
25 



106. Timber Columns. — The allowable fiber stress on columns exceed- 
ing 10 diameters in length may be obtained by the formula 

C xl 



P = C 



f C Xl \ 
Vioo X d) 



100 X d t 

where C is the unit stress given above for short columns; / is the length 
and b the least width of the column, both in inches. 



Details or Construction 

107. General. — Adjustable members will be, if possible, avoided in all 
parts of the structure. 

108. Symmetrical Sections. — All sections shall preferably be made 
symmetrical. 

109. Minimum Number of Rivets. — Excepting lattice bars all connec- 
tions shall have at least two rivets. 

no. Minimum Thickness of Material. — Excepting for lining or fillers 
no material shall be less than £ inch thick. 



372 GRAPHICS AND STRUCTURAL DESIGN 

Note. — Some designers place this minimum thickness at -fe inch for material 
protected from the weather and not affected by injurious gases. Where the con- 
ditions might be expected to cause rapid deterioration of the material it is fre- 
quently made in excess of the \ inch first stated. 

in. Connections. — Connections shall be made strong enough to 
develop the full strength of the members. 

112. Floor Beams. — Floor beams shall ordinarily be rolled sections. 

113. Trusses. — Trusses shall preferably be riveted structures. Heavy 
trusses of long span may be designed as pin-connected structures. 

Roof trusses shall preferably permit of placing purlins at panel points 
only. Where this would prove uneconomical the upper chord may be 
designed to resist combined flexure and compression and the purlins then 
placed where convenient. 

114. Bracing. — Lateral, longitudinal and transverse bracing in all 
structures is preferably composed of rigid members, and should be designed 
of sufficient strength to withstand the wind pressure and any other forces 
acting on it both during and after erection. 

Trusses should be braced in pairs in the plane of their lower and upper 
chords. 

115. Column Splices. — Column splices should be designed to resist the 
bending as well as the direct stresses. 

116. Intersecting Members. — Intersecting members should be located 
so that the axes through their centers of gravity shall intersect in a point. 

117. Separators. — When rolled beams are placed side by side to act 
as a single girder they shall be secured together by bolts and separators 
at intervals not exceeding 5 feet. Beams of greater depth than 10 inches 
should have two bolts in each separator. Parallel beams carrying arched 
floors, whether of the ordinary or flat arch type, shall have tie bolts spaced 
not farther apart than 8 times the depth of the beams. These tie bolts 
should be placed well below the centers of the beams. 

118. Width of Flange Plates. — Flange plates shall be limited in width 
so as not to extend more than 6 inches beyond the line of rivets securing 
the flange plates to the flange angles, nor should this dimension exceed 8 
times the thickness of the thinnest flange plate. 

119. Web Splices. — All web splices must have a plate on each side of 
the web, and these plates must be able to transmit through their rivets the 
full stress coming on the splice at that section. 

Riveting 

120. Riveting. — The minimum distance between centers of rivet holes 
shall be three diameters of the rivet; it is preferred, however, that this 
distance should be not less than 3 inches for f-inch rivets, 2\ inches for 
f-inch rivets, 2 inches for f-inch rivets, and if inches for f-inch rivets. 



STEEL MILL BUILDINGS 373 

The maximum pitch in the line of the stress for members composed of 
plates and shapes will be 6 inches for f-inch and f-inch rivets, 4! inches for 
f-inch rivets and 4 inches for |-inch rivets. 

121. Rivet Spacing in Angles. — For angles built into sections with two 
gauge lines, the rivets being staggered, the maximum pitch in each line shall 
be twice that given in paragraph 120. 

122. Riveting Plates. — Where two or more plates are in contact, they 
shall be held together by rivets whose distance apart in either direction does 
not exceed 12 inches. 

The pitch of rivets in the direction of the stress shall not exceed 6 inches, 
nor 16 times the thickness of the thinnest outside plate. 

The spacing at right angles to the stress shall not exceed 50 times the 
thickness of the thinnest plate. (Some specifications make this 40 instead 
of 50.) 

123. Rivet Spacing from Edges. — The minimum distance of a rivet 
from a sheared edge shall be i| inches for |-inch rivets, i| inches for f-inch 
rivets, if inches for f-inch rivets, and 1 inch for |-inch rivets. The mini- 
mum distances from rolled edges shall be ij, if, 1 and f inch, respectively. 

The maximum distance of a rivet from an edge shall be 8 times the thick- 
ness of a plate. 

124. Calculations of Rivet Strength. — All calculations of rivet values 
both in shear and in bearing shall be based upon their nominal diameters. 

The grips of rivets should preferably not exceed 4 diameters of the rivets. 

Note. — The grip is the total thickness of the pieces held by the rivet. This 
limitation is made to insure the rivet completely filling the rivet hole when driven. 

125. Maximum Rivet Diameters in Angles. — In main members the 
diameters of rivets in angles shall not exceed one-quarter the widths of the 
legs in which they are driven. In other places these diameters may be 
increased f inch. 

126. Pitch of Rivets at Ends of Compression Pieces. — The pitch of 
rivets at the ends of compression pieces shall not exceed 4 times the rivet 
diameter for a distance equal to if times the maximum width of the 
member. 

127. Tie Plates. — Latticed sides of compression members shall have tie 
plates as close to the ends as practicable, and at all intermediate points at 
which the lacing may be omitted. The lengths of these tie plates shall at 
least equal the distance between the rivet lines securing them to the flanges. 
The intermediate plates may be but one-half this length. The thickness 
of these tie plates must be at least tu of the distance between the lines of 
rivets securing the plate to the flanges. 



374 GRAPHICS AND STRUCTURAL DESIGN 

128. Lattice Bars. — Lattice bars shall be proportioned to resist the 
shear corresponding to the allowance for bending made in the column 
formula, paragraph 90. The minimum thickness of lattice bars shall be 
■£0 of their lengths for single lattice bars and wo of their lengths for double 
lattice bars riveted at their centers. 

The minimum widths of lattice bars shall be as follows: 

For 15-inch channels, or built sections 

with 3^ and 4-inch angles, 2\ inches (|-inch rivets). 

For 12, 10 and 9-inch channels, or built 

sections with 3-inch angles, 2j inches (f-inch rivets). 

For 8 and 7-inch channels, or built 

sections with 2^-inch angles, 2 inches (f-inch rivets). 

For 6 and 5-inch channels, or built 

sections with 2-inch angles, if inches Q-inch rivets). 

Note. — The specifications of the American Bridge Co. require the lattice bars 
to be designed to carry a shear of 2 per cent of the direct stress on the column. 

The inclination of the lattice bars with the axis of the member shall not 
ordinarily be under 45 degrees. When the distance between the rivet lines 
in the flange is more than 15 inches, if a single riveted bar is used the lattic- 
ing shall be double, and the bars riveted at their intersection. 

Lattice bars with two rivets shall generally be used in flanges wider than 
5 inches. 

129. Pitch of Lattice Connections along the Member. — The ratio of 
the pitch of lattice connections along a flange to the least radius of gyration 
of that side of the member shall not exceed the ratio of length to least 
radius of gyration for the whole column. 

130. Joints. — In general all joints in riveted work, whether for tension 
or compression members, shall be fully spliced. Joints in compression 
members when the abutting faces are finished for bearing may be spliced 
sufficiently to hold the connecting members accurately in place. 

131. Pins. — A pin shall have a diameter not under f of the width of 
the widest bar held by it. Pins must be turned true to size and straight; 
they must be driven with pilot nuts. 

132. Pinholes. — Pinholes shall be reinforced by plates where neces- 
sary. At least one plate shall be as wide as the projecting flanges will 
allow; where angles are used, this plate shall be on the same side as the 
angles. The plates must contain sufficient rivets that their portion of the 
pin pressure may be distributed to the full cross section of the member. 

Pins must be sufficiently long to insure a full bearing of all parts connected 
upon the turned-down body of the pin. Members should be packed on 
pins to produce the least bending moment on the pin. Vacant spaces 



GIRDER 375 

along the pin should be filled and the members held against lateral move- 
ment. 

133. Temperature Range. — Expansion and contraction shall be pro- 
vided for corresponding to a temperature range of 150 degrees Fahrenheit. 

134. Expansion Rollers. — The minimum diameter of expansion rollers 
shall be 4 inches. 

135. Anchor Bolts. — Columns, when resisting tensile stresses at their 
bases, shall be anchored by bolts to the foundations. Anchor bolts shall 
be long enough to secure a weight of masonry at least one and one-half 
times the tension in the anchor. The minimum size of anchor bolts shall 
be 1 1 inches. 

Materials. — See paragraphs 1 to 18 inclusive. 
Workmanship. — See paragraphs 19 to 57 inclusive. 
Inspection. — See paragraphs 58 to 65 inclusive. 

SPECIFICATIONS FOR A DECK-PLATE GIRDER 
RAILWAY BRIDGE 

136. Material. — All material to be rolled steel as specified in para- 
graphs 1 to 10 inclusive. Cast iron or steel castings will be permitted 
only in machinery of movable bridges and in special cases for shoes and 
bearings. 

137. Plate Girders. — Plate girders are recommended for spans from 
20 feet to 100 feet. 

138. Spacing of Girders. — Deck plate girders shall generally be spaced 
6 feet 6 inches. 

139. Floor. — The floor shall consist of 8-inch by 8-inch cross ties 
separated 6 inches. They shall be notched to fit on the flanges on which 
they should have a full and even bearing. 

140. Guard Rails. — Guard timbers 6 by 8 inches shall be placed on 
each side of the track. Their inner faces shall be not less than 3 feet 3 
inches from the center of the track. These guard timbers shall be notched 
one inch over each tie and shall be fastened to every third tie and at each 
splice by a f-inch bolt. Splices shall be over floor timbers with half-and- 
half joints of 6 inches lap. The floor and guard timbers must be continued 
over piers and abutments. 

Loads 

141. Dead Load. — In estimating the weight of the structure, for use 
in calculating the strains, the timber shall be assumed as weighing 4! pounds 
per foot B. M. (board measure, 144 cubic inches). The weight of the rails, 
spikes and joints shall be taken at 160 pounds per lineal foot of track. 

142. Live Load. — All bridges shall be designed to carry, in addition to 
their own weight and that of the floor, a moving load on each track con- 



376 GRAPHICS AND STRUCTURAL DESIGN 

sisting of two engines coupled at the head of a uniformly distributed train 
load, placed so as to give the greatest strain in the structure. 

This loading is frequently specified Cooper's E-40 as a minimum. 

Note. — The loading Cooper's E-60 is given on page 72. For the other Cooper's 
loadings the distances between the wheels remain the same as the E-60 loading. 
The driving-wheel loads are 1000 times the number designating the loading; thus 
for E-60 the load on the drivers is 60,000 pounds, while for E-40 it is 40,000 pounds. 
The pilot-wheel load is assumed 50 per cent of the load on the drivers, while the 
tender-wheel loads are 65 per cent of the driving-wheel loads. 

143. Impact Allowance. — The effect of impact and vibration shall be 
added to the above-mentioned maximum live-load stresses, and shall be 
determined by the formula, 

V300 -I- L) 
where / = impact to be added to the live-load stresses; 
S = computed live-load stress; 

L = length of loaded distance which produces the maximum stress 
in the member. 
Impact shall not be added to stresses produced by longitudinal, cen- 
trifugal and lateral wind forces. 

144. Wind and Lateral Loading. — All bridges shall be designed for a 
lateral force on the loaded chord of 200 pounds per lineal foot plus 10 per 
cent of the specified train load on one track, and 200 pounds per lineal 
foot on the unloaded chord, these forces being considered as moving. 

Note. — This loading provides for the wind load and vibration and impact due 
to lateral swaying of the train caused by unbalancing of the locomotives, etc. 

Unit Stresses 

145. Unit Stresses in Structural Steel. — All parts of the structure shall 
be so designed that on structural and rivet steels the sums of the maximum 
stresses shall not exceed the following unit stresses, excepting as specified 
in paragraph 91. 



Character of stress. 


Allowable unit stress, 
lbs. per sq. in. 


Axial tension, net section 


16,000 


Flexural stress, extreme fiber stress, net section 

Flexural stress, on pins, extreme fibers 

Axial compression, gross section 


16,000 
24,000 

5=l6,000— (70- J- 



5 = allowable unit stress; 
/ = unsupported length of member, in inches; 
r = least radius of gyration of the member, in inches. 



GIRDER 



377 



146. Stresses on Rivets. — The following are the allowable working 
stresses in pounds per square inch on bolts and rivets in shear and bearing. 







Unit stress. 




Shear. 


Bearing. 


Hand-driven field rivets and turned bolts 


9,000 
11,000 
12,000 
10,000 


18,000 


Power-driven field rivets 


2 2,000 


Shop-driven rivets and pins 


24,000 


Plate girder webs, gross section 




Expansion rollers, per linear inch, where d = 
diameter of the roller in inches 


the 


600 ~X.d 







147. Pressure on Foundations. — The pressures on masonry piers shall 
not exceed the following loads in pounds per square inch. 



First-class sandstone or limestone masonry, and Portland 
cement concrete, 1-2-4, including impact 

First-class granite masonry, and Portland cement mortar, 
including impact 



Pounds per 
sq. in. 



400 
600 



See also paragraph 79. 

148. Pressure on Soils. — See paragraph 77. 

149. Combined Stresses. — See paragraph 93. 

150. Girder Design. — The depth shall preferably be not less than rV 
of the span. When made shallower the flange stresses should be reduced 
so that the deflection will not exceed that of a girder having its depth rs 
of the span. Girders may be designed either by calculating the moments 
of inertia of their net sections or by assuming the flange areas as concen- 
trated at their centers of gravity; in this case | of the gross web section may 
be considered as flange area; when this is assumed the web splices must be 
designed for carrying bending as well as shear. 

151. Flange Design. — The flanges shall be designed so that the area of 
the cover plates shall not exceed 60 per cent of the total flange area. See 
also paragraph 118. 

152. Compression Flange. — The area of the compression flange must 
at least equal the area of the tension flange at the same section of the girder. 
Where the unsupported distance along the girder exceeds 16 times the width 
of the flange the fiber stress in the compression flange must be limited by 



378 GRAPHICS AND STRUCTURAL DESIGN 

that determined by the following formulae. Where the flange consists of 
plates, / = 16,000 — I 200 X - ) ; where the flange is a channel section, / = 

16,000 — ( 150 X 7 ] ; here 

/ is the allowable fiber stress in pounds per square inch; 

/ is the distance between lateral supports along the girder, in inches; 

b is the width of the flange in inches. 

153. Flange Rivets. — The rivets securing the upper flange angles to 
the web plate must be calculated to resist the change in horizontal shear, 
together with such vertical shear due to wheel concentrations and floor 
loads as may be transferred to the web through the flange angles. 

When the rails are placed directly on the upper flange the wheel loads 
shall be assumed as distributed over 30 inches; if ties transfer this load to 
the girder, the load may then be considered as distributed over three ties. 

154. Web Design. — The web shall be designed for the total maximum 
shear,' including dead load, live load and impact; the fiber stress on the gross 
section shall not exceed 10,000 pounds per square inch. See also para- 
graph 100. 

155- Web Splices. — See paragraph 150. 

156. Web Stiff eners. — There shall be web stiff eners, generally in pairs, 
over bearings, at points of concentrated loading, and at other points where 
the thickness of the web is less than eV of the unsupported distance between 
flange angles. The distance between stiff eners shall not exceed that given 
by the following formula, with a maximum limit of six feet and not greater 

than the depth of the web; d = — (12,000 — s). 

40 

d = the clear distance between stiff eners, in inches; 

t = the web thickness, in inches ; 

s = the unit shearing stress in the web, in pounds per square inch. 

157. Bracing. — Lateral bracing shall be placed as close as possible to 
the plane of the upper chord when clearing the ties. 

Spans exceeding 70 feet shall have lateral bracing near the plane of both 
top and bottom chords. Cross frames shall be placed at the ends and at 
intermediate points separated not more than 20 feet. 

158. Minimum Sizes of Materials. — Excepting for fillers the minimum 
thickness of material shall be f inch. Rivets shall not be under £ inch 
diameter. 

159. Rivet Spacing. — See paragraphs 120 to 126. 

160. Expansion. — See paragraphs 133 and 134. 



CONCRETE 379 

All spans exceeding 80 feet in length shall have hinged bolsters at both 
ends, and at one end turned rollers, running between planed surfaces. 

Rollers shall not be under 4 inches in diameter. Bridges with spans 
under 80 feet shall have one end free to move upon planed surfaces. 

161. Camber. — The camber shall be tV inch for every 10 feet. 

Note. — Camber is an upward curvature given the bridge and should theoreti- 
cally be equal and opposite to that caused by the deflection of the bridge under the 
load for which it was designed. It, therefore, serves to bring the nominally hori- 
zontal members into actually horizontal positions when subjected to their full load. 
Under full load then the several members will be in the same relative positions as 
those for which their stresses were determined. 

SPECIFICATIONS FOR PORTLAND CEMENT CONCRETE 
AND REINFORCED CONCRETE 

162. Cement. — Cement shall be Portland, and shall meet the require- 
ments of the standard specifications of the American Society for Testing 
Materials. 

163. Fine Aggregates. — These shall consist of sand, crushed stone, 
or gravel screenings graded from fine to coarse, and when dry passing a 
screen having holes \ inch in diameter; it shall preferably be of siliceous 
material, clean, coarse, free from vegetable loam or other deleterious matter, 
and not more than 6 per cent shall pass a sieve having 100 meshes per linear 
inch. 

164. Test. — Mortars composed of one part Portland cement and three 
parts fine aggregate by weight when made into briquettes shall show a 
tensile strength of at least 70 per cent of the strength of 1-3 mortar of 
the same consistency made with the same cement and standard Ottawa 
sand. 

165. Coarse Aggregates. — These shall consist of crushed stone or 
gravel, graded in size, which is retained on a screen having holes \ inch 
in diameter; it shall be clean, hard, durable and free from all deleterious 
matter. Aggregates containing soft, flat or elongated particles shall not be 
used. 

166. Maximum Size of Coarse Aggregate. — The maximum size of the 
coarse aggregate shall be such that it will not separate from the mortar in 
laying and will not prevent the concrete fully surrounding the reinforce- 
ment or filling all parts of the forms. 

Where concrete is used in mass the maximum size of the coarse aggregate 
may, at the option of the engineer, be such as to pass a 3 -inch ring. For 
reinforced concrete, sizes usually are not to exceed one inch in any direction, 
but may be varied to suit the character of the reinforcement. 



38o 



GRAPHICS AND STRUCTURAL DESIGN 



167. Water. — Water used in mixing concrete shall be free from oil, 
acid, alkalies, or vegetable matter. 

Reinforcing Steel 

168. Metal Reinforcement. — The metal reinforcing steel shall be manu- 
factured from new billets and shall meet the requirements of the following 
specifications and be free from rust, scale, or coatings of any character 
which would tend to reduce or destroy the bond. 

169. Process of Manufacture. — See paragraph 1. 

170. Chemical and Physical Requirements. — 



Requirements . 


Structural steel. 


High carbon 
steel. 


t,, , ( Basic 


0.04% 
0.06% 
0.05% 

( Desired 
( 60,000 
1,500,000 


0.085% 
0.075% 


Phosphorus, max. < * c [a 


Sulphur, maximum 


Ultimate tensile strength, lbs. per sq. in.. . 

Elongation, min. per cent in 8 ins., Fig. 1 . . . 

Elongation, min. per cent in 2 ins., Fig. 2. . . 
Character of fracture 


Desired 

85,000 

1,400,000 


Ult. ten. str. 
22% 
Silky 
180 flat 


Ult. ten. str. 




Cold bends without fracture 


180 d = 4t 







171. Yield Point. — The yield point, as indicated by the drop of the beam, 
shall be not less than 60 per cent of the ultimate strength. 

172. Allowable Variations. — See paragraph 3. 

173. Chemical Analyses. — See paragraph 9. 

174. Forms of Specimens. — See paragraph 10. 

175. Number of Tests. — At least one tensile and one bending test shall 
be made from each melt of steel as rolled. 

176. Modifications in Elongation. — See paragraph 4. 

177. Bending Tests. — Bending tests may be made by pressure or blows. 
Shapes and bars less than one inch thick shall bend as called for in para- 
graph 170. 

178. Bending Tests of Thick Material. — See paragraph 5. 

179. Finish. — See paragraph 14. 

180. Stamping. — See paragraph 15. 

181. Defective Material. — See paragraph 16. 

Concrete 

182. Proportions. — The materials to be used in concrete shall be of 
uniform quality and so proportioned as to secure as nearly as possible a 
maximum density. 



CONCRETE 381 

183. Unit of Measure. — The unit of measure shall be the barrel, which 
shall be taken to be 3.8 cubic feet. Four bags containing 94 pounds of 
cement each shall be considered the equivalent of one barrel. Fine and 
coarse aggregates shall be measured separately as loosely thrown into the 
measuring receptacle. 

184. Relation of Fine and Coarse Aggregate. — The fine and coarse 
aggregate shall be used in such relative proportions as will insure maximum 
density. 

185. Relation of Cement and Aggregates. — For reinforced-concrete 
construction a density proportion based on one to six shall be used, i.e., 
one part of cement to a total of six parts of fine and coarse aggregates 
measured separately. 

186. Mixture for Massive Concrete. — For massive or rubble concrete 
a density proportion based on one to nine shall be used. 

187. Mixing. — The ingredients of concrete shall be thoroughly mixed 
to the desired consistency, and the mixing shall continue until the cement 
is uniformly distributed and the mass is uniform in color and homogeneous. 

188. Measuring Proportions. — Methods of measurement of the pro- 
portions of the various ingredients, including the water, shall be used, which 
will secure separate uniform measurements at all times. 

189. Machine Mixing. — When the conditions will permit a batch mixer 
of a type which insures uniform mixing of the materials throughout the mass 
shall be used. 

190. Hand Mixing. — When it is necessary to mix by hand the mixing 
shall be on a watertight platform and especial precautions shall be taken 
to turn the materials until they are homogeneous in appearance and color. 

(a) Tight platforms shall be provided of sufficient size to accommodate 
men and materials for the progressive and rapid mixing of at least two 
batches of concrete at the same time. Batches shall not exceed one cubic 
yard each, and smaller batches are preferable, based upon a multiple of 
the number of sacks of cement to the barrel. 

(b) Spread the fine aggregates evenly upon the platform, and the cement 
upon the fine aggregates; mix these thoroughly until of an even color. 
Add all the water necessary to make a thin mortar and spread again; add 
the coarse aggregates, which, if dry, should first be thoroughly wet down. 
Turn the mass with shovels or hoes until thoroughly incorporated and all 
the aggregates are thoroughly covered with mortar; this will probably 
require the mass to be turned four times. 

(c) Another approved method, which may be permitted at the option of 
the engineer in charge, is to spread the fine aggregates, then the cement, 
and mix dry, then the coarse aggregates; add water and mix thoroughly 
as above. 



382 GRAPHICS AND STRUCTURAL DESIGN 

191. Water. — The materials shall be mixed wet enough to produce a 
concrete of such a consistency as will flow into the forms and into the metal 
reinforcement, and which, on the other hand, can be conveyed from the 
place of mixing to the forms without separation of the coarse aggregate from 
the mortar. 

192. Retempering. — Remixing concrete with water after it has partially 
set will not be permitted. 

193. Placing Concrete. — (a) After the addition of water, it shall be 
handled rapidly from the place of mixing to the place of final deposit, and 
under no circumstances shall concrete be used that has partially set before 
final placing. 

(b) The concrete shall be deposited in such a manner as will permit the 
most thorough compacting, such as can be obtained by working with a 
straight shovel, or slicing tool kept moving up and down until all the in- 
gredients have settled in their proper place by gravity and the surplus water 
has been forced to the top. 

(c) In depositing concrete under water, special care shall be exercised 
to prevent the cement from floating away, and to prevent the formation 
of laitance. 

Note. — Laitance is a whitish gelatinous substance, of about the same composi- 
tion as cement but having little or no hardening properties. It is caused by the 
action of water on the surface of the concrete. 

(d) Before placing concrete the forms shall be thoroughly wetted and the 
space to be occupied by the concrete freed from debris. 

(e) When work is resumed, concrete previously placed shall be roughened, 
thoroughly cleaned of foreign material and laitance, drenched and slushed 
with a mortar consisting of one part of Portland cement and not more than 
two parts of fine aggregate. 

(f) The faces of concrete exposed to premature drying shall be kept wet 
for a period of at least seven days. 

194. Freezing Weather. — The concrete shall not be mixed or deposited 
at a freezing temperature, unless special precautions, approved by the engi- 
neer, are taken to avoid the use of materials containing frost or covered 
with ice crystals, and to provide means to prevent the concrete from freez- 
ing after being placed in position and until it is thoroughly hardened. 

195. Rubble Concrete. — Where the concrete is to be deposited in 
massive work, clean stones, thoroughly embedded in the concrete as near 
together as is possible and still entirely surrounded by concrete, may be 
used at the option of the engineer. 

196. Forms. — Forms shall be substantial and unyielding and built so 
that the concrete shall conform to the designed dimensions and contours, 



CONCRETE 383 

and so constructed as to prevent the leakage of mortar. These forms shall 
not be removed until authorized by the engineer. 

197. Forms for Important Work. — For important work, the lumber 
used for face work shall be dressed to a uniform thickness and width, and 
shall be sound and free from loose knots, secured to the studding or up- 
rights in horizontal lines. 

198. Less Important Work. — For backings and other rough work 
undressed lumber may be used. 

199. Round Corners. — Where corners of masonry and other projections 
liable to injury occur, suitable moldings shall be placed in the angles of 
the forms to round or bevel them off. 

200. Re-using Lumber. — Lumber once used in forms shall be cleaned 
before using again. 

201. Wetting Forms. — In dry but not freezing weather the forms shall 
be drenched with water before the concrete is placed against them. 

Details of Construction 

202. Splicing Reinforcement. — Whenever it is necessary to splice the 
reinforcement by lapping, the length of lap will be decided by the engineer 
on the basis of the safe bond stress in the reinforcement at the point of 
splice. Splices shall not be made at the points of maximum stress. 

. 203. Joints in Concrete. — Concrete structures, wherever possible, shall 
be cast in one operation, but when this is not possible the work shall be 
stopped, so that the resulting joint shall have the least effect on the strength 
of the structure. 

204. Placing Girders and Slabs. — Girders or slabs shall not be placed 
over freshly formed walls or columns without permitting a period of at least 
two hours to elapse to provide for settlement or shrinkage in the supports. 
Before resuming work the top of the supports should be thoroughly cleansed 
of foreign matter and laitance. 

205. Temperature Changes. — In massive work, such as retaining walls, 
abutments, etc., built without reinforcement, joints shall be provided 
approximately every 50 feet throughout the length of the structure to care 
for the temperature changes. To provide against the structures being 
thrown out of line by unequal settlement, each section of the wall may 
be tongued and grooved into the adjoining section. To provide against 
unsightly cracks, due to unequal settlement, a joint shall be made at sharp 
angles. 

206. Surface Finish. — The desired finish of the surface shall be deter- 
mined by the engineer before the concrete is placed, and the work shall be 
so conducted as to make it possible to secure the finish desired. 



384 GRAPHICS AND STRUCTURAL DESIGN 

Plastering of surfaces will not be permitted. 

Note. — The preceding portion of this specification is taken from those of the 
American Railway Engineering and Maintenance of Way Association, Proceedings 
of 1909. 

207. Fireproofing. — For ordinary conditions, it is recommended that 
the metal in columns and girders be protected by a minimum of two inches 
of concrete; that the metal in beams be protected by one and one-half 
inches of concrete and that the metal in slabs be protected by at least one 
inch of concrete. In monolithic columns a section around the column 
having a width of at least one and one-half inches should be allowed for fire- 
proofing and not estimated as carrying any load. All corners of columns 
should be rounded, as rounded corners are affected less by fire and are also 
less liable to other injury. 

Loads 

208. Vertical Loads. — The same vertical loads as those given under 
" Steel Mill Building Specifications " may be followed; see paragraphs 
68, 69, '70 and 74. Buildings in cities should be designed in accordance with 
the local building laws. Where such laws are not available the following 
floor loads may be assumed: 

Office floors 75 lbs. per sq. ft. 

Floors for light running machinery 75"-i5° lbs. per sq. ft. 

Floors for medium heavy running 

machinery 200 lbs. per sq. ft. 

Storage, to be estimated from weight of 

the materials in each case 150-1000 lbs. per sq. ft. 

The weight of the concrete shall be assumed as 150 pounds per cubic foot. 
For Loads on Bridges see paragraphs 141, 142 and 143. 

209. Wind Loads. — These will be the same as those given under the 
previous specifications; see paragraphs 71, 72, 73 and 144. 

210. Impact. — Impact may be taken into account by making additions 
to the live-load stresses. 

an. Dimensions for Calculations. — (a) The spans of beams and girders 
shall be assumed as the distance center to center of supports but need not 
be taken greater than the clear span plus the depth of the beam or girder. 

(b) For slabs supported at the ends the span shall be assumed as the 
clear span plus the depth of the slab. 

(c) For continuous slabs the spans shall be taken as the distances between 
the centers of supports. 

212. Bending Moments. — (a) For continuous beams and slabs the 
bending moments at the center and at the supports shall be assumed as 



CONCRETE 



385 



80 per cent of the moment on a freely supported beam having the same span 
and load. 

Note. — The Specifications of the Joint Committee of the A. S. C. E., A. S. T. M., 
A. R. E. and M. of W. Assn., etc., give this as 67 per cent for interior spans but 
80 per cent for end spans. In the case of beams and slabs continuous for two 
spans only, the bending moment at central support shall be calculated, using 100 
per cent as the factor, while near the middle of the span the factor shall be assumed 
80 per cent. 

(b) Floor slabs should be designed and reinforced as continuous over 
the supports. If the length of the slab exceeds one and one-half times its 
width, the entire load shall be carried by the transverse reinforcement. 

Square slabs shall be reinforced in both directions. 

Note. — The distribution of loads in the two directions upon slabs having a 
ratio of length to width not exceeding 1 to 1.5 will be approximately as given in 
the following table. 



1 

b' 


r. 


I 

b' 


r. 


1 .0 


0.50 


i-3 


0-75 


1 .1 


o-59 


i-4 


O.80 


1 .2 


0.67 


1-5 


0.83 



Here 

r is the proportion of the load carried by the transverse reinforcement, 

/ is the length, and 

b is the breadth of the slab. 
Using the values just specified, each set of reinforcements is to be calculated 
in the same manner as for slabs having supports on two sides only, but the 
total amount of reinforcement thus determined may be reduced 25 per cent 
by gradually increasing the rod spacing from the third point to the edge 
of the slab. 

213. T Beams. — In beam-and-slab construction, an effective bond 
should be provided at the junction of the beam and slab. When the prin- 
cipal slab reinforcement is parallel to the beam, transverse reinforcement 
should be used, extending over the beam and well into the slab. 

Where adequate bond and shearing resistance between the slab and web 
of beam is provided, the slab may be considered as an integral part of the 
beam, but its effective width shall be determined by the following rules: 
(a) It shall not exceed one- fourth of the span length of the beam; (b) its 
overhanging width on either side of the web shall not exceed four times the 
thickness of the slab. 



386 GRAPHICS AND STRUCTURAL DESIGN 

In the design of T beams acting as continuous beams, careful considera- 
tion must be given to the compressive stresses' at the supports. 

Beams in which the T form is used only for the purpose of providing 
additional compression area of concrete should preferably have a width of 
flange not more than three times the width of the stem and a thickness of 
the flange not less than one- third of the depth of the beam. 

Both in this form and in the beam and slab form, the web stresses and 
the limitations in placing the longitudinal reinforcements will probably be 
controlling factors in the design. 

214. Internal Stresses. — The internal stresses in slabs, beams and 
girders shall be determined by the formulae recommended by the Joint 
Committee of the A. S. C. E., etc., which are given in the body of the book, 
page 215. Throughout the entire beam the shearing and bonding stresses 
must be determined and proper provision made to develop the required 
strength. 

215. Columns. — The ratio of the unsupported length of a column 
divided by its least width shall not exceed 15. The effective area of a 
column shall be considered as the area within the fireproofing, or in the case 
of hooped columns, or columns reinforced with structural shapes, it shall 
be taken as the area within the hooping or structural shapes. 

The following working fiber stresses may be allowed in columns varying 
with the class of reinforcement in it. 

. (a) Columns with longitudinal reinforcement only, such reinforcement 
not less than 1 per cent nor more than 4 per cent. The unit stress shall be 
those allowed for axial compression in Working Stresses, paragraph 218. 

(b) Columns with reinforcements of bands, hoops or spirals, as hereafter 
specified: the stresses allowed shall be 20 per cent higher than those given 
for (a), provided the ratio of the unsupported length of the column to 
the diameter of the hooped core is not more than eight. 

(c) Columns reinforced with not less than 1 per cent and not more than 
4 per cent of longitudinal bars, and with hoops or spirals, as hereafter 
specified: the stresses shall be 45 per cent higher than those given for (a), 
provided the ratio of the unsupported length of the column to the diameter 
of the hooped core is not more than eight. 

Note. — In all cases the longitudinal reinforcement is supposed to carry its pro- 
portion of the stress. The hoops or bands are not to be counted on as. adding 
directly to the strength of the column. 

Bars composing longitudinal reinforcement shall be straight and shall have suffi- 
cient lateral support to be securely held in place until the concrete has set. 

Where hooping is used, the total amount of such reinforcement shall not be less 
than 1 per cent of the volume of the column inclosed. The clear spacing of such 
hooping shall not be greater than one-sixth of the diameter of the inclosed column, 



CONCRETE 



387 



and preferably not greater than one-tenth; in no case shall it exceed 2.5 inches. 
The hooping must be circular, and the ends of bands must be united to develop 
the full strength of the bands. Adequate means must be provided to hold the 
hoops or bands in place so that the column shall have a straight and well-centered 
core. As the effect of hooping decreases rapidly with the increase of ratio of length 
to core diameter the above fiber stresses are limited to columns in which this ratio 
does not exceed 8. 

216. Bending Stresses on Columns. — Bending stresses due to eccentric 
loading and lateral forces must be provided for by increasing the section 
until the maximum stress does not exceed the values above specified; and 
where tension is possible in the longitudinal bars, adequate connection 
between the ends of the bars must be provided to take this tension. 

217. Reinforcing for Shrinkage and Temperature Stresses. — Rein- 
forcement not under one-third of 1 per cent, and able to develop a high 
bonding strength shall be provided; it shall be placed near the exposed 
surfaces and be well distributed. 



Working Stresses 

218. Working Fiber Stresses. — The following working stresses for 
concrete are based on the compressive strength developed by the concrete 
after 28 days, when tested in cylinders 8 inches in diameter and 16 inches 
long. Such tests should show the following ultimate strengths, in pounds 
per square inch. 



Aggregate. 



Granite, trap rock 

Gravel, hard limestone and hard 

sandstone 

Soft limestone and sandstone ...... 

Cinders 



1:1:2 


1:1.5:3 


1:2:4 


1 : 2.5 : 5 


3300 


2800 


2200 


1800 


3000 


2500 


2000 


1600 


2200 


1800 


1500 


1200 


800 


700 


600 


500 



1:3:6 



1400 

1300 

IOOO 
400 



(a) Bearing. — When compression is applied to a surface of concrete of 
at least twice the loaded area, a stress of 32.5 per cent of the ultimate com- 
pressive strength may be allowed. 

(b) Axial Compression. — For concentric compression on a plain con- 
crete column or pier, the length of which does not exceed 12 diameters, 
22.5 per cent of the ultimate compressive strength of the concrete may be 
allowed. See also paragraph 215. 

(c) Compression in Extreme Fibers. — The extreme fiber stress on a 
beam, calculated on the assumption of a constant modulus of elasticity for 
the concrete under working stresses, may reach 32.5 per cent of the ultimate 



3 88 



GRAPHICS AND STRUCTURAL DESIGN 



compressive strength of the concrete. Adjacent to the supports of con- 
tinuous beams, these stresses may be increased 15 per cent. 

(d) Shear and Diagonal Tension. — In calculations on beams in which 
the maximum shearing stress in a section is used as the means of measur- 
ing the resistance to diagonal tension stress, the following allowable values 
for the maximum vertical shearing stresses are recommended. 

1. For beams with horizontal bars only and without web reinforcement, 
2 per cent of the ultimate compressive strength may be allowed. 

2. For beams thoroughly reinforced with web reinforcement, the value 
of the shearing stress having been calculated, using the total vertical shear 
in determining the unit horizontal shear, the working shearing stress should 
not exceed 6 per cent of the ultimate compressive strength of the concrete. 
The web reinforcements, exclusive of the bent-up bars in this case, shall 
be proportioned to resist two-thirds of the external vertical shear. 

3. For beams in which part of the longitudinal reinforcement is used 
in the form of bent-up bars distributed over a portion of the beam in a way 
covering the requirements of this type of web reinforcement: the limit of 
the allowable working shearing stress shall be 3 per cent of the ultimate 
compressive strength of the concrete. 

4. Where punching shear occurs, that is, shearing stress uncombined 
with compression normal to the shearing surface, and with all tension normal 
to the shearing plane provided for by reinforcement; a shearing stress of 
6 per cent of the ultimate compressive strength of the concrete may be 
allowed. 

(e) Bond. — The bond stress between plain reinforcing bars and concrete 
may be assumed as 4 per cent of the ultimate compressive strength of the 
concrete. In the case of drawn wire, 2 per cent should be the limit. 

219. Steel. — The working fiber stress in the steel shall not exceed 
16,000 pounds per square inch. 

220. Modulus of Elasticity. — The ratio of the moduli of elasticity of 
steel and concrete will vary with the ultimate compressive strength of the 
concrete. The following values may be used. 



Strength of concrete, lbs. per sq. in. 


Ratio of moduli 

of elasticity, 

concrete to 

steel. 


2200 and less 


1 


Exceeding 2200 but under 2900 


Exceeding 2900 





CHAPTER XXII 
PROBLEMS 

The object of these problems is to furnish work for the student paralleling 
that in the drawing room. Introductory review problems in Mechanics 
are given that lead up to the principal problems in the book. As far as pos- 
sible, all the problems have been made similar to those occurring frequently 
in practice. Although the tables of sections in the book give sufficient data 
for the working of the problems the writer believes that the small additional 
outlay for one of the handbooks issued by the manufacturers of structural 
steel is money well spent. Among those issuing these books are the Cambria 
Steel Company of Johnstown, Pa., and the Carnegie Steel Company and 
Jones & Laughlin Company of Pittsburgh, Pa. 

Center of Gravity 
i. Prove that the distance from the base to the center of gravity of a 
rectangle is \ h. 

2. Prove that the distance from the base to the center of gravity of a 
triangle is I h. 

3. Prove that the distance from the base to the center of gravity of a 
half-round section is c = 0.4244 R. 

Note. — Use the fact established in Problem 1 and make the infinitesimal 
strips run from the diameter to the circumference. 








m" 


"V 

i 

i 

. i. 






f 




i 




! 

8" 




1 ' 

— ! 


1 

i 



Fig. 1. 



Fig. 2. 



Fig. 3. 



4. Prove that in the given trapezoid, Fig. 1, 



_i 3 b + b 



3 2 b -f &i 
389 



x Xh. 



39° 



GRAPHICS AND STRUCTURAL DESIGN 



5. Calculate the distance from the back to the center of gravity of a 
15-in. channel weighing 33 lbs. per ft. See p. 16. 

6. In a 6 X 4 Xf-in. angle calculate the distance of the center of gravity 
from the back of the short leg. 

7. How far from the base is the center of gravity in Fig. 2? 

8. How far from the base is the center of gravity in Fig. 3? 

9. Prove that the center of gravity of a semicircular arc is c = — 

IT 

from the diameter joining its extremities. 

10. How far from its base is the center of gravity of a cone? How far 
from its base is the center of gravity of its lateral surface? 

Moments of Inertia 

1= (yHA. 

Given I the moment of inertia, then I' referred to any axis parallel to 
the principal axis is /' = I -\- Ah 2 . 

In these problems do not recalculate positions of centers of gravity. 

11. Calculate the moment of inertia about the principal axis of a rec- 
tangle of altitude h and width b. 



■h 



1 - 


2 

nl 


*_i 


' 




1 



Fig. 4. 



2 
Fig. 5. 




Fig. 6. 



12. Calculate the moment of inertia of a triangle about an axis through 
its axis center of gravity and parallel to its base. 

13. Calculate the moment of inertia of a circle for an axis through its 
center. (Use polar coordinates.) 

14. Calculate the moment of inertia of a 6 X 4 X I -in. angle referred to 
an axis through its center of gravity and parallel to its short leg. 

15. Calculate the inertia of a 12-in. channel at 20^ lbs. per ft. for an axis 
through its center of gravity and parallel to its back. 

16. Calculate the inertia of two 10-in. channels at 15 lbs. per ft. referred 
to axis 2-2 when the distance between channel backs is 6.33 ins. (See Fig. 4.) 

17. Calculate the inertia about axis 2-2 for two 5 X 3 X |-in. angles 
placed as in Fig. 5. 



PROBLEMS 



391 



18. Calculate the inertia for the horizontal axis passing through center of 
gravity of Fig. 2. 

19. Calculate the inertia for axis 1-1 of four 4 X 4 X Hn. angles placed 
as in Fig. 6. 

20. Calculate the inertia of a half-round for axis 1-1, Fig. 7. 



k— r 




IT 



T 



t 



Fig. 7. 



Fig. 8. 



— L— 
Fig. 9. 



1 



21. Calculate the inertia of the girder section referred to axis 1-1, Fig. 8. 
Web J-in. plate. 
Angles 4 X 4 X f in. 
Flange plates 10 X T 7 6 m - 
Assume the section contains two f-in. diam. holes for f-in. rivets through 
horizontal legs of angles and flange plates for which allowance must be made. 



Reactions and Bending Moments 

It should be recalled that for equilibrium when the forces act on a section 
or a point that the 

Sum of the horizontal forces = o. 
Sum of the vertical forces = o. 
Sum of the moments of forces = o. 

The vertical shear at a section is the algebraic sum of the vertical forces 
to the left of that section. 

22. A cantilever beam of length L ft. carries the load P at its free end. 
What is the reaction at the support, bending moment at this point and also 
bending moment a distance X from free end? 

23. A cantilever beam of length L ft. carries a uniform load of W lbs. 
per ft. What is the reaction, the bending moment at the support and also 
at a distance X ft. from the free end? 

24. A cantilever beam carries a load of W lbs., varying uniformly from 
zero at the free end to a maximum at the support. Required the reaction, 
the bending moment at the support and at a distance X ft. from the free end. 

25. A simple beam of span L ft. carries a central load of W lbs. What 
are the reactions, the maximum bending moment and the bending moment 
X ft. to the right of the left support? 



39 2 



GRAPHICS AND STRUCTURAL DESIGN 



26. A simple beam carries a uniform load of W lbs. per ft. Its span is L ft. 
What are the reactions, the maximum bending moment and the bending 
moment X ft. to the right of the left support? 

27. A simple beam, Fig. 9, carries a load W, varying from zero at sup- 
ports to a maximum at the middle. Required the reactions, bending moment 
at the middle and at a distance X ft. from the left support. 

28. A simple beam, Fig. 10, carries a load W lbs., varying from zero 
at the left support to a maximum at the right support. Required the reac- 
tions, the maximum bending moment and the distance it occurs from the 
left support. 



l^WtTTTTJ 



§ 



I^TTHlflU M^-frj fTFT, 



Fig. 10. 



Fig. 11. 



Fig. 12. 



29. Required the reactions, maximum bending moment, and bending 
moment X ft. from the left support in Fig. 11. 

30. In Fig. 1 2 find the reactions and maximum bending moment. What 
is the bending moment 1 5 f t. from the left support? 

31. In Fig. 13 what are the reactions? What is the bending moment 
over support? What is the bending moment X ft. to the right of the left 
support? What is the maximum bending moment and where does it occur? 



!*-C-^*— 



oi 



-(£*» 



Fig. 13. 



k20^ 



f 15 ' 1 



— w ! — -^4 

Fig. 14. 



C.L. 



Fig. 15. 






=T 



32. The given beam in Fig. 14 in addition to the loads shown carries 
a uniform load of 500 lbs. per ft. of span. What is the maximum bending 
moment and where does it occur? What is the bending moment 28 ft. 
from the left support? 

33. Prove that in Fig. 15 when two loads P, a constant distance a apart, 
roll across a girder the maximum bending occurs when a load is a distance 

- from the center of the span. 



PROBLEMS 393 

Selection of Beam Sections 

In selecting beams from manufacturers' handbooks it is common practice 
to use a factor called the section modulus which is tabulated in the hand- 
books. The section modulus = -or moment of inertia divided by the 

e 

distance from the neutral axis to the extreme fibers. 

Since M = •*— > it follows — - = - ; 

e f e 

hence, first find — by dividing the maximum bending moment in inch 

pounds by the allowable working fiber stress. In the following problems 
assume the beams sufficiently stiffened laterally. Problems where the ratio 
of span to flange width must be considered will be taken up later. 

34. Select a beam to carry a uniform load of 1500 lbs. per lineal foot on a 
span of 20 ft.; allow a working fiber stress of 15,000 lbs. per sq. in. 

35. A beam of 32-ft. span carries four loads of 5000 lbs. each, spaced 8 ft. 
apart, while the first is 4 ft. from the left support. Allow 15,000 lbs. fiber 
stress per sq. in. and select an economical beam section, neglecting the 
bending due to the beam's weight. 



hl0#l2#i 3 4 § I I I 



4 



6'[ 



—50* H l*«sl*— 24- — *W ^8^12^124^ * 25 ' *i 

Fig. 16. Fig. 17. Fig. 18. Fig. 19. 

36. In Fig. 16 assume also a uniform load including weight of beam as 
100 lbs. per lineal foot. Select an economical beam section, allowing a 
working fiber stress of 12,000 lbs. per sq. in. 

37. In Fig. 17 neglect weight of beam, allow 12,000 lbs. fiber stress and 
select a suitable section. 

38. The loading is as shown in Fig. 18. Neglect weight of beam. Allow 
a working fiber stress of 12,000 lbs. per sq. in. and select an economical 
beam section. 

39. Select I beam for Fig. 19. Assume beam secured laterally. The 
moving wheel loads are 6 ft. o ins. c-c. Each wheel load is 10,000 lbs. 
Use a working fiber stress of 12,000 lbs. per sq. in. Assume dead load 
including weight of beam as 60 lbs. per ft. 

40. Select I beam for the span in Fig. 20. Assume beam secured later- 
ally. The moving wheel loads are 12 ft. o ins. c-c. Each wheel load is 
1 2 ,000 lbs. Allowable working fiber stress is 1 2 ,000 lbs. per sq. in. Assume 
dead load including weight of beam as 80 lbs. per ft. 



394 ) GRAPHICS AND STRUCTURAL DESIGN 

Deflection of Beams 

It is not proposed to review the entire discussion of deflection of beams 
but simply to recall the subject by one or two of the easier cases and then 
solve a few problems by the use of the formulae given on page 7. The 
general equation of the elastic curve for all beams is 

&y _ M, 
dx 2 ~ EI' 

here M = bending moment due to the external forces at a section whose 
abscissa is X. 
I = moment of inertia of beam section. 
E = modulus of elasticity of the material. 
y — ordinate along which deflection is measured. 







f — ^-—A 

Fig. 20. 

41. A cantilever beam, Fig. 21, of uniform section carries a concentrated 
load W at its free end. What is its deflection if its length is Z? 

Assume origin of coordinates at o. Here M = WX and the general 
equation becomes 

EIp-=-WX. 
ax 2 

Integrate this twice and eliminate the constants of integration by means 

of the following facts. At the fixed end, -—•= o and X = L, also Y = o 

ax 

when X = o. 

42. Prove that the deflection of a simple beam of span L and central 

load W is 

WD 

4 8£/' 

Find the form of the equation of the elastic curve for this beam, Fig. 22, 
and loading. Integrate twice, remembering that 

-f- — o when X = - and that X = o when y = o. 
ax 2 

43. A standard 15-in. I beam 42 lbs. per ft. having a span of 30 ft. is to 
carry a uniform load: its working fiber stress is not to exceed 16,000 lbs. 



PROBLEMS 395 

per sq. in. and the maximum deflection — - — of its span. What load will it 

360 

carry? 

44. A 6-in. I beam 12^ lbs. per ft. extends 5 ft. from a wall as a cantilever 
beam. Assume that it is amply stiff, laterally. What load will it carry 
if the working fiber stress is 12,000 lbs. per sq. in. and what will be its 
maximum deflection? 

45. What uniform load will a 12-in. I beam 31! lbs. per ft. carry on a 
20-ft. span if it is merely supported at the ends and the fiber stress is 16,000 
lbs. per sq. in.? Compare the deflection and fiber stress of this beam 
with a beam having fixed ends, carrying the same uniform load. 

Note. — In practice one generally assumes simple beams, beams merely 
supported at the reactions. Fixed and continuous beams are seldom de- 
signed. Reinforced concrete designers make allowance for fixing and con- 
tinuity in their beam designs. 

Tension Pieces 

In the usual tension piece of constant cross section the stress is uniformly 
distributed over the cross section and is the same for all sections. The 
section may not be constant, in which case the minimum or net section must 
be considered. 

46. A 2-in. round bar is to have its end upset for U. S. Standard screw 
thread. What size screw must be cut if the area at the root of the thread 
exceeds the area at the body of the rod by 18 per cent? 

47. A 4 X 4 X f-in. angle carries a load of 35,000 lbs. in tension. Assume 
two £f-in. diameter holes in a cross section. What is the unit stress per 
square inch on the net section? 

Columns 

In the case of columns the stress tends to increase any initial flexure due 
to inaccurate workmanship or vibration and a column, when of sufficient 
length, fails by bending. It should be noted that tension in a piece reduces 
any buckling tendency. 

Columns Failing by Flexure 

The derivation of column formulae is based upon the general equation 
of the elastic curve previously used in finding the deflection of beams. In 
these column formulae, of which there are several, a quantity depending 

entirely upon the section of the column occurs; it is— , and its square root 



396 GRAPHICS AND STRUCTURAL DESIGN 

is called the radius of gyration of the section referred to the same axis as /. 
Here / is the moment of inertia and A the area of the section. 



~. ,. , .. 4 /inertia of section * fl 

Radius of gyration = V ~* ~7 : = V t* 

* Area of section ▼ A 

48. Two 8-in. channels iij lbs. per ft. are spaced such a distance back 
to back that the moments of inertia referred to their two principal axes 
are equal. What do the radii of gyration equal? 

49. Two 6 X 4 X i-in. angles are placed with their long legs parallel 
and separated f in. What are the radii of gyration referred to the principal 
axes? 

50. Two s X3I X l-in. angles are placed with their short legs parallel 
and separated f in. What are the radii of gyration referred to the two 
principal axes? 

51. Calculate the radius of gyration of a circular section 8 ins. in diameter. 
Then calculate the radius of gyration of a ring section 8 ins. outside diam- 
eter and'6 ins. inside diameter. 

52. Prove the following values of radii of gyration: 

Rectangle, principal axis r = 



v> 



Triangle, principal axis r = — ■— . 

V18 



Circle, principal axis r = -■ 

4 



The following formulae are frequently used for mild-steel column design: 

Ends. Fixed. Hinged. 

_, , . , -, l6,000 ,, l6,000 

Rankine s / = , 7 = — 



P 

1+ -= : 1 + 



10,000 r 

Straight line/' = 16,000 — 35 -, /' = 16,000 — 70 - 

r T 

The most rational formula for columns is that of Ritter, 

/ 



/ = 



1 + 

mir 2 E 



fe*ey 



PROBLEMS 397 

/' = allowable unit working stress, pounds per square inch. 
/ = length of column, inches. 
r = least radius of gyration. 
/ = maximum desired fiber stress, occurring at the dangerous section of 

the column and resulting from the flexure of the column. 
f e = fiber stress per square inch at the elastic limit of material. 
E = coefficient of elasticity of the material of the column. 
m = a constant, depending upon the way the column ends are secured. 
Its value for the usual end conditions are : 



Ends. Both Fixed. i Fixed and i Hinged. 


Both Hinged. 


m 4 2.25 


I 


For soft or mild steel, taking E = 30,000,000, 




it 2 = 10 approximately 
and 




f e = 30,000; 




the formula becomes, for hinged ends, 




,, 16,000 




* t /A 2' 




1 + -*- ( l - ) 

10,000 \r 1 





53. Compare the allowable working stresses as given by the three for- 
mulae for a mild-steel hinged column, whose length is 20 ft. and least radius 
of gyration of the column section is r = 2. 

54. Compare the allowable working stresses as given by the three for- 
mulae for a mild-steel fixed-end column whose length is 20 ft. and least radius 
of gyration of the column section is r = 2. 

55. Two 15-in. channels weighing S3 lbs. per ft. are to be made into a 
latticed column. Assume hinged ends, use Ritter's formula and determine 
the maximum load which they can be designed to carry if the column is 
35 ft. long. 

56. What load will a Bethlehem 12-in. H section weighing 78 lbs. per ft. 
carry? Assume the column, with fixed ends, 25 ft. long and the material 
mild steel. Use Ritter's formula and compare the result with that given 
in the Bethlehem Steel Company's handbook, using the formula 

f = 16,000- 55-. 

57. Compare the loads that can be carried by two mild steel columns, 
each 22 ft. long, both having fixed ends, the first being a latticed column 
made of two 10-in. channels 15 lbs. per ft., spaced for maximum load, the 



398 GRAPHICS AND STRUCTURAL DESIGN 

other being a io-in. column 54 lbs. per ft., Bethlehem Steel Company's 
section. Use Ritter's formula in both cases. 

In structural design it is usual practice to limit the value of - for impor- 

r 

tant work subjected to considerable shock to - = 100, while for other work 

r 

the usual limit is- = 120. 
r 

58. Would a 6 X 6 X f-in. angle used singly make a good column? 
Should it be used for a height of 16 ft.? Why? What load could it carry 
if 10 ft. long, with hinged ends? Use Ritter's formula and assume the 
material mild steel. 

59. Would an 8-in. I beam weighing 18 lbs. per ft. make a good column? 
Why? Should it be used for a height of 12 ft.? Why? What load would 
it carry if 8 ft. high? Use Ritter's formula. Assume fixed ends and the 
material mild steel. 

60. Two 6 X 4 X f-in. angles are to be placed back to back, separated 
I in. and used for a hinged strut. Take your data from Manufacturer's 
Handbook and show if it would be more economical to have the long or 
short legs parallel. What load would it carry if 12 ft. long? Use Ritter's 
formula and assume the material mild steel. 

In the beams in the previous problems no consideration has been taken 
of their lateral stiffness . In practice it is customary when using a maximum 
working fiber stress to limit the ratio of unsupported beam length to flange 
width to from 12 to 20. Where the ratio exceeds these numbers the allow- 
able working fiber stress should be reduced. This is made necessary by the 
fact that in a vertically loaded beam the upper flange, being subjected to 
compression, is liable to fail as a column by buckling and, therefore, to 
secure the same factor of safety, the working stress in this flange must be 
reduced. There are several ways by which allowance is made for this condi- 
tion, all of which are empirical and subject to criticism. Attention is called 
to the reduction of the working fiber stress by the use of a column formula, 
as in the Cambria Handbook, 

r __ l8,000 
J ~~ T) 



1 + 



3000 b L 

} = allowable stress in pounds per square inch. 
/ = length between lateral supports in inches. 
b = width of flange in inches. 



PROBLEMS 399 

This formula is for a maximum desired working stress of 16,000. For 
limiting values other than 16,000 lbs. reduce the maximum fiber stresses by 
the percentage that the above formula reduces 16,000 lbs. 

Another method is based upon the tests of long beams by the Pencoyd 
Iron Works and adopted by other companies' handbooks. It assumes the 
maximum allowable fiber stress applicable to spans of 20 flange widths 
and a uniform reduction as the ratio of span to flange width increases 
until a beam whose span equals 70 flange widths is reached when the 
allowable working fiber stress is limited to one-half that used in the 
first case. 

Still another method is that explained in Chapter VII of this book. 

61. What uniform load will a 15-in. I beam weighing 42 lbs. per ft. carry 
on a 23-ft. span if unsupported laterally, and if the maximum working fiber 
stress for a beam whose span is 20 flange widths is limited to 16,000 lbs. 
per sq. in.? Compare the results by the methods given above. 

62. What central load can be carried upon a Bethlehem girder beam G-20 
weighing 112 lbs. per ft. on a 30-ft. span? The maximum fiber stress for 
a span of 20 widths is 16,000 lbs. per sq. in. Compare the results obtained 
by the methods given. 

* 63. A 10-ton crane is to be carried across a span of 30 ft. on rolled 
beams. The maximum wheel load is 21,000 lbs., the wheel base 10 ft. 6 ins. 
The maximum desired working stress is 15,000 lbs. per sq. in. reduced to 
allow for ratio of span to flange width. Select standard Cambria sections 
and test their ability to resist a lateral pull of one-tenth of the live load 
divided between the two wheels acting on the beam. 

* 64. A 25-ton crane is to be carried across a 36-ft. span on rolled beams. 
The maximum wheel load is 40,000 lbs.; the wheel base is 10 ft. 6 ins. All 
other conditions are the same as in Problem 63, excepting that a Bethlehem 
Grey Mill section with a wide flange is to be used. Determine the section. 



Web Stresses 

In addition to the points already considered, in designing beams with 
heavy concentrations of loads or very short spans the ability of the web to 
resist vertical shearing, vertical crippling and horizontal shearing must also 
be taken into account. In the case of plate girders the web is reinforced 
with stiffening angles to resist this crippling. Rolled steel beams are rarely 
weak in vertical or horizontal shear; timber beams should be carefully 
examined for horizontal shear; while in reinforced-concrete construction 

* Specifications frequently call for an impact allowance of 25 per cent in the 
case of traveling cranes. 



400 GRAPHICS AND STRUCTURAL DESIGN 

careful designing to resist the web stresses, including horizontal shear and 
diagonal tension, is of vital importance. 
The horizontal shear at any section of a homogeneous beam is 

f s = horizontal shear on section per square inch. 

/ = moment of inertia of entire section. 

V = total vertical shear. 

w = width of web. 
Say = statical moment of the area of the cross section on one side of the 
neutral axis, or summation of areas multiplied by their re- 
spective distances from the neutral axis. 

65. Show that for a timber beam of width b and depth d 

U 2bd 

66. If a 15-in. I beam weighing 42 lbs. per ft. is to be used on varying 
spans and a fiber stress of 16,000 lbs. is allowed in tension and 12,000 lbs. 
in shear, what is the minimum span for which the horizontal shear may be 
neglected? 

67. Allowing a tensile fiber stress of 1500 lbs. per sq. in. and a shearing 
fiber stress with the grain of 120 lbs. per sq. in., what uniform load can 
be carried by a yellow pine timber 12 ins. deep, 3 ins. wide on a 12 -ft. span? 

68. Allowing a tensile fiber stress of 900 lbs. per sq. in. and a shearing 
fiber stress with the grain of 100 lbs. per sq. in., what uniform load will 
a spruce timber 16 ins. deep X 4 ins. wide carry on a 16-ft. span? 

69. What is the maximum span upon which a yellow pine timber 12 ins. 
deep can be used to carry a uniform load if its fiber stress is to be limited 
to 1200 lbs. per sq. in. and its deflection to -gio of its span? Assume its 
modulus of elasticity as 1,500,000 lbs. per sq. in. 

70. What central load will a white oak timber 16 ins. deep X 4 ins. wide 
carry on an 18-ft. span if stressed to 1500 lbs. per sq. in.? What would 
be its maximum deflection if the modulus of elasticity were taken at 
1,500,000 lbs. per sq. in.? 

Combined Flexure and Direct Stress 
Sometimes beams are subjected to direct compression or tension in addi- 
tion to flexural stresses. It is usually sufficient to take the algebraic sum 
of the flexural and direct stresses at the extreme fibers of the beam. When 
greater accuracy than this is required, Johnson's formula, as modified by 



PROBLEMS 401 

Merriman, can be used. According to this formula the fiber stress due to 
bending, when direct compression or tension acts upon a piece in addition 
to bending, is 

f = Me 

m E 
The — is for a compression force. 
The + is for a tensile force. 

P 

The combined stress is/ ± — • 

P = compression or tensile force in pounds. 
/ = span in inches. 

Simple beam uniform load = — = — • 

m 9.6 

Simple beam central load= — = — 

m 12 

A = area of cross section in square inches. 

An exact method is given by Merriman but its application is too difficult 
to be generally used. 

71. A yellow pine beam 10 ft. long and 12 ins. square is subjected to a 
compression force acting along its length of 50,000 lbs. while carrying a 
uniform load of 20,000 lbs. What is the fiber stress? E = 1,500,000. 

What is the maximum fiber stress, using the formula 

,, = Me 



m E 



72. A horizontal steel tension bar has a cross section of 9 X i£ ins. 
It is subjected to a unit stress in tension of 10,000 lbs. per sq. in. The bar 
is 20 ft. long. Determine the maximum resulting fiber stress due to the 
combined action of its weight and its tensile load. The bar runs horizontally 
with its i|-in. edge down. E = 30,000,000 

n 1 

m 9.6 

73. In Fig. 23 the top plate is 20 X f in., the sides are 15-lb. channels 
at 40 lbs. per foot each. Assume the gross cross section loaded with 7500 



4-02 



GRAPHICS AND STRUCTURAL DESIGN 



lbs. per sq. in. Chord 25 ft. long. E = 3o,ooo;ooo. Determine the 
maximum fiber stress due to direct loading and its own weight. Use formula 



f 



Me 



/T 



nPP 
m E 



74. In Fig. 24 the top plate is 24 X 1 in. The sides are, two upper 
angles 4 X 4 X T 9 g in., two plates 24 X ^f in. and two bottom angles 
6 X 4 X 1 in. Area of combined section is 82.3 sq. ins. Inertia of sec- 
tion is 6969. Assume chord 20 ft. long. Uniform load, including live load 



Fig. 23. 



J 



L 



Fig. 24. 



Fig. 



25- 



and weight of section, is 4000 lbs. per foot of span. Compression load 
400,000 lbs. acting along the center of gravity of the section. What is the 
total maximum fiber stress? E = 30,000,000. 



/' = 



Me> 



I - 



nPP 
m E 



1 
9.6' 



75. What is the maximum fiber stress in Fig. 25? Two angles 4 X 
3 X T 5 6 m -> one plate 10 X T Vin. Compression due to direct load acting 
at center of gravity of section is 10,000 lbs. per sq. in. on gross section. 
(In this problem neglect rivet holes.) Length of piece 8 ft. o ins. Neg- 
lect weight and assume a central load due to purlin of 1600 lbs. What is 
the maximum fiber stress? 

Me 



/' = 



/ ± i? 



E = 30,000,000. 



PROBLEMS 403 

Riveting 

The following considerations are useful in designing riveted joints for 
structural work. 

1. The rivet strength is calculated upon the nominal diameter of the 
rivet notwithstanding the fact that the hole in which the rivet is driven 
is generally rs in. greater than the rivet. 

2. The unit shearing strength of steel rivets is taken about £ the unit 
tensile strength of the steel and the unit bearing strength of rivets is taken 
at double their unit shearing strength. 

3. Rivets should not be used in tension. 

4. Use table of rivet values. 

76. A 24-in. I beam weighing 80 lbs. per ft. carries a uniform load on a 
19-ft. span. First find what load it will carry if the extreme fiber stress is 
16,000 lbs. per sq. in. and then find how many f-in. rivets will be required 
to secure two 4 X 4 X f-in. angles to the web of the beam and transfer 
the reaction due to the given load. Compare your result with the stand- 
ard framing. Allow 10,000 lbs. per sq. in. shear. 

77. Two 4 X 4 X xVin. angles are fastened back to back to a f-in. plate. 
How many f-in. rivets will be required in the angles if the net section 
of the material is stressed 12,000 lbs. per sq. in.? Allow one rivet hole 
f in. diam. through angles and plate. Allow 7500 lbs. per sq. in. in shear. 

78. The long leg of a 5 X 3I X f-in. angle is fastened to a f-in. plate. 
Assume one rivet hole f in. in diameter, in a section. Allow a fiber stress 
of 12,000 lbs. per sq. in. on the net section and determine how many f-in. 
rivets are required. Allow 7500 lbs. per sq. in. in shear. 

Timber Columns 

The U. S. Dept. of Agriculture has made a large number of tests upon 
timbers and the following is their formula suggested for timber columns. 

p = px 7°o+ i SC 

700+I5C + C 2 

To find the safe load divide p by the required factor, where 

p = ultimate strength in pounds per square inch. 
F = crushing strength of timber. 

_ Z, Length (inches) 

d small diameter (inches) 

79. What load can be safely carried by a yellow pine post 10 ins. square 
and 20 ft. high? Use a factor of 4 and consider the ultimate crushing 
strength 5000 lbs. per sq. in. 



404 GRAPHICS AND STRUCTURAL DESIGN 

80. A cedar post is 22 ft. high and is 10 X 12 ins. in cross section. What 
load can it safely carry if its ultimate crushing strength is assumed at 3500 
lbs. and a factor of safety of 5 is desired? 

81. A spruce column is 18 ft. high and 9 ins. square. What load can it 
safely carry if* its ultimate crushing strength per sq. in. is taken as 4000 lbs. 
and a factor of safety of 4 is desired? 

Plate Girders 

In designing girders of greater depth than rolled sections (24 and 36 ins.) 
the following method is easier and is sufficiently accurate. In Fig. 26, 
A = area of one flange in square inches. 

h = distance between the centers of gravity of the two flanges in 
inches. Where the flange is made up of angles and plates 
this distance is frequently assumed as the distance back to 
back of the angles. 
/ = mean fiber stress in the flange in pounds per square inch. 
M = external bending on the section measured in inch pounds. 

Then M = A XfXh. 

This formula assumes the web as resisting shear only. When the web is 
assumed to resist bending also the formula becomes 

M = I A + fl X f X h. 



-(' + !)*> 



Here a is the area of the web plate in square inches. 

82. Derive the two formulae for girders, 

M = AxfXh and M = (A + -\xfXh. 

83. Find the net flange area required at the middle of the following girder. 
Span 60 ft. Depth of girder back to back of flange angles is 6 ft. o ins. 
Uniform dead load is 600 lbs. per ft. (this covers ties, rails and metal of girder, 
etc.). Uniform live load is 2250 lbs. per lineal foot. Increase the live-load 
bending 80 per cent to allow for impact due to a moving train load. Allow a 
fiber stress of 16,000 lbs. per sq. in. Assume that the web takes shear only. 

84. A girder spans 80 ft. Its depth, back to back of flange angles, is 7 ft. 
o ins. Uniform dead load is 700 lbs. per ft. (this covers ties, rails, metal of 
girder, etc.). Uniform live load is 2000 lbs. per lineal foot. Increase the 
live-load bending 80 per cent to allow for impact due to a moving train 
load. Allow a fiber stress of 16,000 lbs. per sq. in. Assume that one- 
eighth of the web acts as flange area and that the web is f in. thick. What 



PROBLEMS 



405 



additional area is required to complete the net flange section 30 ft. from the 
left abutment? 

85. A girder which spans 50 ft. is to be made 5 ft. b ins. deep, back to 
back of flange angles. It carries two wheel loads 12 ft. o ins. center to 
center of 80,000 lbs. each. Assume a dead load including girder weight of 
200 lbs. per ft. Increase the live-load bending 20 per cent to cover impact. 
Allow a fiber stress of 16,000 lbs. per sq. in. Assume the web re in. thick 
and that one-eighth of it is considered as being flange. Determine what 
additional area is required to complete the net flange area at the point of 
maximum bending. 



Tr r 



JLi 



Fig. 26. 



U 22- --H 



Fig. 27. 




Fig. 28. 



86. A bridge which spans 60 ft. is to be built for a crane. Assume that 
each girder weighs 15,000 lbs. and that this forms a uniform load. The 
trolley wheels are 6 ft. apart and each wheel load is 16,000 lbs. The girder 
at the center is 48 ins. deep, back to back of angles. On account of stiffen- 
ing the crane laterally the flange plates are made 22 ins. wide. Assume the 
section like Fig. 27. Assume the web as taking shear only. The fiber stress 
in the compression flange is to be 9000 lbs., while that in the tension flange 
can be 12,000 lbs. per sq. in. Determine the dimensions of the plates. 

Shafting 
Considering the differential area, Fig. 28, the fiber stress is proportional 
to its distance r from the center so that if p is the fiber stress at a distance 
R from the center the stress p r at this distance is 



Pr = 



P Xr 



The force acting on an area dA then is dF = p r XdA. 
The moment of this force about the center then is 



Integrating 



dM t 



M 



pXr 2 
R 



dA. 



= |/*u. 



406 - GRAPHICS AND STRUCTURAL DESIGN 

It will be noticed that this is in the form of the general equation for 
torsion. 

M t = ^, 

where / is the polar moment of inertia. 

87. Allowing a limiting shearing working stress of 7000 lbs. per sq. in. 
upon a steel round 2 ins. in diameter, what twisting moment measured in 
inch pounds will the round carry? What would be the force applied at a 
radius of 1 ft.? 

88. Allowing a limiting shearing working stress of 7000 lbs. per sq. in., 
what twisting moment in inch pounds would be carried by a 2 in. square 
shaft? Acting at a radius of 1 ft. what force would this correspond to? 

89. Find the ratio of the twisting moments carried by a round shaft 
and a square shaft of the same sectional area and the same limiting fiber 
stress. 

90. What horse power will a shaft of diameter d transmit when making 
N revolutions per minute and subjected to a working fiber stress of p lbs. 
per sq. in. at its circumference? 

Torsional Deflection 

It is sometimes desirable to limit the torsional deflection of a shaft, in 
which case the angle of deflection in degrees is given by 

A= M t xL 
1660 X d* 
M t = twisting moment in inch pounds. 
L = length of shaft in feet. 
d = diameter of shaft in inches. 

91. A shaft 2 ins. in diameter and 60 ft. long is subjected to a twisting 
moment which produces an extreme fiber shearing stress of 6000 lbs. per 
sq. in. Through what angle is one end of the shaft twisted ahead of the 
other ? 

Combined Bending and Twisting 

Shafting must generally be designed for combined torsion and bending. 
There are several formulae but according to Guest's law the following may 

be used for mild steel. 

M e = VAP + T 1 . 

Here M e — equivalent bending moment. 

M = bending moment. 
T = twisting moment. 



PROBLEMS 



407 



Having found the equivalent bending moment, the shaft diameter can be 
determined by placing these values in the equation 

jr.-**- 

e 

I = moment of inertia referred to axis. 

e = distance from center of gravity to extreme fibers. 

92. In Fig. 29 a force of 2000 lbs. acts on the teeth of a pinion 5 ins. in 
diameter. The distance from the center line of the pinion to the center of 
the adjacent bearing is 6 ins. What diameter of shaft will be required if 
9000 lbs. per sq. in. is permitted in flexure? 




Fig. 31 



Fig. 32. 



93. In Fig. 30 a force of 2500 lbs. acts upon the teeth of a pinion \\ ins. 
in diameter. The pinion is inside the bearings as shown. Using a nexural 
fiber stress of 9000 lbs. per sq. in. what diameter of shaft is required? 



Stresses in Structures Determined Algebraically 

In Fig. 31 to determine the stress in any piece, as FA, cut the truss at a 
section containing this piece and then equate the internal and external 
moments about any point that most conveniently gives the desired result. 
Consult Chapter IV. 

In Fig. 32 BC = 2400 lbs., CD = DE = EF = etc. = 4800 lbs. 

94. Calculate algebraically the stress in GC and AG in Fig. 32. 

95. Determine algebraically the stresses in GH and HD in Fig. 32. 

96. Calculate the stresses in HI and I A in Fig. 32. 

97. Calculate the stresses in FL and LM in Fig. 32. 
Problems 94 to 97 should .be checked graphically. 

In Fig. ss BC = CD = DE = EE' , etc. = 31,250 lbs. 

98. Calculate algebraically the stresses in BF and FA in Fig. 33. 

99. Calculate algebraically the stresses in FG and GC in Fig. 33. 
100. Calculate algebraically the stresses in GH and HI in Fig. 33. 
Problems 98 to 100 should be checked by method of coefficients. 



4o8 



GRAPHICS AND STRUCTURAL DESIGN 



Graphical Analysis of Truss Stresses 

ioi. In using the common graphical method of determining truss stresses, 
what assumption is made concerning the construction of the truss? What 
are the three conditions necessary for equilibrium in a structure acted on by- 
forces? 




^dObk 



Fig. 34. 



102. What is a force triangle? A force polygon? What can you say 
concerning the force polygon representing a number of concurrent forces 
in equilibrium? What general direction will the forces in the polygon take? 

103. Make a sketch of any simple truss and show how force polygons 
may be used to determine the magnitude and character of the stresses in the 
structures when the external forces are known. 

104. Take Fig. 34, letter the truss and make a careful free-hand sketch 
of the stress diagram, then find the character of the stresses. 




X 



^ c 



i_l 



1 



Fig. 35. 



1 F 

Fig. 36. Fig. 37. 



105. Take Fig. 35, letter the truss and make a careful free-hand sketch 
of the stress diagrams, first with a live load L at 1 and then at 2. Deter- 
mine the character of the stress. 

106. What is an equilibrium polygon? When a number of noncon- 
current forces are in equilibrium what graphical conditions are fulfilled? 

107. In Fig. 36 show by a free-hand sketch how the resultant of the 
three forces acting on the piece can be determined in direction, magnitude 
and located in relation to the piece. 

108. In Fig. 37 show by a free-hand sketch how the reactions due to the 
three parallel forces acting on the beam can be determined. 

109. In Fig. 38 a truss is acted on by wind forces. Both ends of the 
truss are fixed. Show how the reactions can be found. 

no. In Fig. 39 the truss is acted on by wind pressure. The right-hand 
end is on rollers while the other end is fixed. Show by a free-hand sketch 



PROBLEMS 



409 



how the magnitude of the right reaction and the magnitude and direction 
of the left reaction can be found. 

in. Explain how an equilibrium polygon may be used as a bending- 
moment diagram and prove that the statement is correct. 

112. Make a free-hand sketch showing by an equilibrium polygon how 
the bending can be found on the beam in Fig. 40. 




Fig. 38. 



Fig. 39. 



Fig. 40. 



The equilibrium polygon may also be used in determining the stresses as 
follows: In finding truss stresses algebraically the internal and external 
moments were equated. The number of calculations can be reduced by 
using an equilibrium polygon for the external bending moment, the single 
diagram serving for all points. 

113. In Fig. 41 show how the stresses AI, HI and HE may be deter- 
mined by combining both graphical and algebraic methods as suggested. 

114. In Fig. 42 determine the dead-load stresses in members AJ, JK 
and IF. The bridge has the following dimensions: Span, 150 ft.; height, 
center to center chords, 30 ft.; uniform dead load, 2000 lbs. per ft., carried 
at lower apex points. 







A 






/ 


X 


K 


/ 


\ 


f.l 


E > 


, D , 


• C N 


, B ' 



Fig. 41. 




Fig. 42. 



Fig. 43. 



115. Show how a bending-moment diagram may be determined graphi- 
cally to give the maximum moments on all sections of a girder for two loads, 
a constant distance a apart, moving across the girder. 

116. Show how a bending-moment diagram may be determined graphi- 
cally to give the maximum moments on all sections of a girder, due to a 
number of loads (use 6), a constant distance apart, moving across the girder. 

117. Show how to construct a diagram giving the maximum shears at 
all points on a girder due to a locomotive and train load passing across it. 
Prove that the diagram gives the required results. 



4io 



GRAPHICS AND STRUCTURAL DESIGN 



118. In Fig. 43 show how the maximum stresses due to moving wheel 
loads crossing the bridge may be determined for members AH, HI and //. 
Use a combined graphical and algebraic method. 

Stresses in Crane Frames 

Show how to determine the stresses in the following crane frames. Ex- 
plain the methods fully. 

i. Find direct stresses due to live load at maximum radius. 

2. Find direct stresses due to live load at minimum radius. 

3. Draw bending-moment diagrams for members subjected to bending. 
If necessary draw a stress diagram for position of live load producing 
maximum bending in any member. 

4. Draw stress diagram for dead load. 




Fig. 44. 



Fig. 45- 



Fig. 46. 



119. Make the necessary stress and moment diagrams for the frame given 
in Fig. 44. 

120. Make the necessary stress and moment diagrams for the frame given 
in Fig. 45. 

I2i. Make the necessary stress and moment diagrams for the frame given 
in Fig. 46. 



"t—^-7^ TTT"^ 1 





Fig. 47. 



Fig. 48. 



122. Make the necessary stress and moment diagrams for the frame 
given in Fig. 47, taking the chain or rope pull into account. 



PROBLEMS 411 

123. Make the necessary stress and moment diagrams for the frame 
given in Fig. 48, taking the chain or rope pull into account. 

124. In Fig. 48 show how to stiffen FB so that the carriage can travel 
back close to the post. Also show how channels used for EA could be 

stiffened to reduce the - value and permit the load chain to travel back 
r 

close to the post. 

Plate-gieder Design 

Consult Chapter XI for information concerning these problems. 

125. Assume a half dozen wheel loads and their distances apart and 
explain how the diagram of maximum bending moments would be drawn. 
Show how to include the bending due to the girder weight in this 
diagram. 

126. A girder has a span of 65 ft. Assume that it carries a load equiva- 
lent to a uniform load of 2000 lbs. per ft. Calculate the bending at the 
center and draw the bending-moment diagram. 

Lengths or Flange Plates 

In plate girders the flange force varies from zero at the supports to a 
maximum near the center of a girder. If the girder is of approximately 
constant depth the bending-moment diagram can represent the flange force 
by changing the scale; thus, 

M = A XpXh 



and the flange force 



F-Ax,-£, 



h being constant F varies as M. This diagram can now be used to deter- 
mine the lengths of the several plates and angles composing the flange. 

127. Assume any girder. Show how to determine the lengths of the 
several flange plates. 

Flange Riveting 

128. For stationary loads the diagram representing the forces acting in 
the girder flanges can be used to determine the riveting of the flange angles 
to the web and also the several flange plates together. Show how to do this. 

129. A girder spans 60 ft. Assume the depth back to back of angles as 
6 ft. The flange at the middle is composed of two 6 X 6 X f-in. angles, 
and two 15 X f-in. plates. The web is | in. thick, rivets are f in. in diameter. 



412 GRAPHICS AND STRUCTURAL DESIGN 

Assume the bending as due to a uniform load on the girder of 6000 lbs. per 
ft. of span. By the method suggested in Problem 128, find the number of 
rivets required between the web and flange angles for the panel extending 
from 6 ft. to 12 ft. from the left pier. Allow 11,000 lbs. shear and 22,000 lbs. 
bearing on the rivets. 

130. Take the same data as in Problem 129, and determine the number 
of rivets in the panel extending from 24 ft. to 30 ft. from the left pier. 

131. Explain how to construct a diagram of maximum shears for a 
locomotive and train load passing across a bridge. 

132. Estimate the reaction from the data given in Problem 129 (this 
includes dead load, live load and impact) and determine the shearing fiber 
stress in the web plate which is 6 ft. deep and f in. thick. 

133. Explain the object of web stiffeners. What 

is the usual spacing of web stiffeners in ordinary ^5 ^Z 
railroad practice ? How thick must the web be, in 
relation to the girder depth, to permit the omission 
of stiffeners ? 



134. Show by means of a diagram of maximum _r ji 






"o-j 


-*4 S 


I+. 


| O <t> 


<i> 



shears that the maximum vertical shear at any 
point due to a moving uniform load occurs when the p IG 4Q 

girder is loaded from this point to the more re- 
mote of the two reactions. 

Instead of the method previously indicated for obtaining the rivet spacing 
between the flange and the web, the following is the method more generally 
used as it is applicable to moving loads. In Fig. 49 

RXh 

s = rivet pitch, inches. 
R = rivet value, pounds. 

h = distance between gauge fines of flange angles, inches. 
V = vertical shear, generally taken as the mean of the panel in 
which the rivets are to be spaced. 

135. Assume the same general data as given in Problem 129 but consider 
the load a moving one. Use the method just given and determine the rivet 
spacing in the lower flange for the section of the girder extending from 12 ft. 
to 18 ft. to the right of the left support. Assume that the web takes shear 
only. Assume that h = 64I. Use £-in. rivets, allowing 11,000 lbs. in shear 
and 22,000 lbs. in bearing. 

136. Given f-in. web, 6 X 4 X re -in. flange angles, h = 55 ins. Rivet 
diameter f in., shearing stress 10,000 lbs. per sq. in., bearing stress 20,000 



PROBLEMS 413 

lbs. per sq. in., vertical shear 75,000 lbs. What rivet spacing is required 
at this section? Assume that the web takes shear only. 

When concentrated loads are transferred to the web plates through the 
flange angles the resultant of the vertical and horizontal forces must be 
found and the rivet spacing determined from this resultant. These con- 
centrations are usually assumed as distributed over a certain distance along 
the flange, say 36 ins. 

The change in the horizontal flange force per inch of flange length is 

The vertical force per inch of flange length is 

load 



h 



distance distributed 

The resultant f r = V/i 2 +/ 2 2 . Rivet spacing = j . 

fr 

137. Solve Problem 129, taking into account the load of 6000 lbs. per foot 
carried from the flange angles to the web by the rivets, h = 64.75 in. 
Use a moving load. 

138. Given web plates f in. Angles 6 X 6 X f in. Rivets | in. in diameter. 
Shearing stress 11,000 lbs. Bearing stress 22,000 lbs. Total vertical shear 
at middle of panel under consideration 200,000 lbs. h = 69 ins. Assume 
load of 45,000 lbs. distributed over 36 ins. of upper flange. (This includes 
dead load, live load and impact.) Determine the rivet spacing. Assume 
that the web takes shear only. 

The horizontal shear between the flange and the web is reduced when the 
web is assumed as assisting in carrying the bending. The change in this 
shear transferred through the rivets depends upon the ratio 

A — a 

where 

A = net area of flange, total. 
a = one-eighth gross web area 



A — a 



fe=flX ^—^ and fr = Vy 3 2 +/2 2 ; 



and, as before, rivet spacing = — ■ 

Jr 



414 



GRAPHICS AND STRUCTURAL DESIGN 



139. In Problem 138 assume that one-eighth of the web acts with the 
flange and that at this point the net flange area is 12 sq. ins., not including 
one-eighth of the web. Find the required rivet spacing. 

140. Determine the rivet spacing satisfying the following conditions: 
Vertical shear 160,000 lbs.; h = 64.75 m s. Net flange area, not including 
one-eighth of web, 20 sq. ins.; web 72 X f in. Upper flange concentration 
45,000 lbs. acting over 36 ins. Shearing stress 11,000 lbs. and bearing 
stress 22,000 lbs. per sq. in. Use rivets f-in. in diameter. 



U 5x3 x£f fi L? 



-%sw.eb 



Fig. 50. 



^Ir^/ 



Fig. 51. 




Fig. 52. 



141. A girder running over two supports is cantilevered 5 ft. beyond one 
of them and carries a concentrated load at its end, producing an average 
fiber stress in the flange angles of 15,000 lbs. per sq. in. Assume the section 
given in Fig. 50 and determine the spacing in the flange angles of the canti- 
levered portion for f-in. rivets. Allow 10,000 lbs. per sq. in. in shear and 
double this in bearing. 

142. Fig. 51 is made up of 1-8 in. LJ 11 1 lbs. per ft., area 3.35 sq. in. 
and 1-12-in. /31'! lbs. per ft., area 9.26 sq. ins. The average upper flange 
stress is 8000 lbs. at the middle of a 22-ft. span, carrying a central load. 
What rivet spacing should be used in securing the channel to the beam if 
the rivets are f in., the shearing fiber stress 10,000 lbs. per sq. in. and 
the bearing stress double this? 

143. In Problem 142 determine the rivet spacing 4 ft. from the left 
support if the beam carries a uniform load instead of a central one. 

Web Splice 



When the web is assumed as resisting bending it becomes necessary to 
rivet the splice plates so that they and the riveting shall replace the broken 
section in bending as well as in shear. See Chapter XL 

144. A girder is 5 ft. deep and has a &-in. web plate; one-eighth of the 
web is assumed as acting as flange; the average flange stress acts 29 ins. from 
the neutral axis and is 12,000 lbs. per sq. in. Using two splice plates, how 



PROBLEMS 



415 



thick must they be made if they can be only 50 ins. deep ? The fiber stress 
in the splice plates is to equal that in the web plate at the same distance 
from the neutral axis of the girder. 

145. Assume the data the same as that given in Problem 144 and Fig. 52. 
How many vertical rows of rivets will be required to make the web splice 
properly carry its portion of the moment ? Allow 10,000 lbs. per sq. in. in 
shear and 20,000 lbs. in bearing. Use | in. diameter rivets. 

146. A girder 6 ft. deep has a f-in. web plate which is to be spliced at the 
middle. How thick must each of two splice plates be if they can be made 
only 58^ ins. deep? The fiber stress at the backs of the angles is 12,000 lbs. 
and the fiber stress in the splice plates is to equal the fiber stress in the cor- 
responding position in the web plates. 

147. Take the girder in Problem 146 and Fig. 53. How many rows of 
rivets will be required, allowing 12,000 lbs. per sq. in. shear and 24,000 lbs. 
in bearing? Design the splice to resist bending. Use £-in. diameter 
rivets. 



R=15000# R 




Fig. 53. 



Fig. 54. 



Fig. 55. 



148. Assume that a girder 6 ft. deep and 60-ft. span carries a movmg 
load of 6000 lbs. per ft. Determine the maximum shear at the middle and 
design the splice plates, assuming that the web takes shear only. How 
many f-in. rivets carrying 12,000 lbs. per sq. in. shear and 24,000 lbs. per 
sq. in. bearing will be required, the web being | in. thick? Depth of splice 
plates is 58 \ ins. 

149. The lateral bracing, Fig. 54, in a deck girder is placed in the plane 
of the upper flanges. Assume that the tension pieces take the load. The 
wind load is assumed as 300 lbs. per ft., to account for wind blowing on the 
train, and 30 lbs. per sq. ft. of girder. The combined load is assumed as 
acting upon the upper flange through the rail. (1) Determine the apex 
loads and (2) make a stress diagram showing how the forces acting in the 
wind braces may be determined. Assume vertical girders 6 ft. deep. 

150. Take the data given in Problem 149 and determine the stresses in 
the bracing algebraically. 



416 



GRAPHICS AND STRUCTURAL DESIGN 



151. Assume that the lateral truss in line with the upper flanges trans- 
mits all the wind forces in this plane to the ends of the girder, and that from 
here the forces are transferred to the piers by the end bracing, Fig. 55. 
Find the reactions from Problem 149. Assume that \ R goes to 2, the other 
£ R through 1 and then to 3. Determine the forces in 1, 2 and 3, and 
design them, assuming struts with fixed ends and that / = § the diagonal. 
Allow a fiber stress of 16,000 lbs. per sq. in. 

Fixed and Continuous Beams 

Fixed and continuous beams do not have much application in structural 
work; however, owing to the monolithic character of concrete work the 
bending moments are frequently taken at values between those for supported 
and fixed beams. 

152. Fig. 56 is a beam fixed at one end and supported at the other; load 
is at the middle of the span. The point of inflection is at X = f /. Show 
how to find the b ending-moment diagram graphically. 



■<-x->\ 

-5 b- 




Fig. 56. 



Fig. 57. 



Fig. 58. 



153. A beam fixed at one end and supported at the other carries a uniform 
load. The reaction at the fixed end is f the load. Determine the bending 
moments graphically and locate the points of inflection. 

154. The beam in Fig. 57 carries a uniform load W lbs. per ft. over 
its length (L+ 2 x). Determine graphically the value of x in terms of L 
so that the central span shall fulfill the conditions of a fixed beam, i.e., 



M 



WU 



24 



at the middle, and at the supports 



M = - 



WD 



155. A fixed beam, Fig. 58, carries a central load W. The bending at 
the supports is — M, while for the middle the bending is + M. (Numeri- 
cally equal but of opposite signs.) Determine graphically these moments 
and the points of inflection. 



PROBLEMS 417 

156. If a beam carries a uniform load and is restrained at the supports 
so that the bending at the middle is T V wl 2 , determine graphically the bend- 
ing moment at the supports and the points of inflection. 

Reinforced-concrete Designing 

For the explanation of the theory and derivation of the formulae see 
Chapter XIV. Also consult page 198 for the nomenclature. 

157. Derive the following formulae: 

M 8 = A Xf 8 Xjd; M c = ^XkjXbXd 2 

2 



and t-S/ ftf+fv 

V b Xfc(3* + 2) 

158. Design a rectangular beam to carry a load of 800 lbs. per ft. of a 

20-ft. span. Assume M = Allow /, = 16,000 lbs., /«. = 500 lbs. 

10 

Take d = 20 ins. and assume — * = 15. Determine the width and the area 

h c 

of the reinforcing steel. 

159. What size rectangular beam and reinforcement will be required 
for a span of 16 ft. The beam ends are merely supported and the load 

is 1000 lbs. per ft. of span. Given M = — — > -=r=. 15, tensile stress in 

8 rL, c 

steel 16,000 lbs. per sq. in., compression in concrete 500 lbs. per sq. in. 

and the width of the beam is to be 40 per cent of its effective depth. 

160. Derive the following approximate formulae for rectangular beams, 
the symbols having the significance given on page 198. 

M 8 = ixAxfsXd, M c =\fcXbXd\ 

161. The purlins on a roof are 6 ft. center to center. How thick must a 
reinforced roof slab be made if the total live and dead load is 80 lbs. per 

sq. ft.? Assume-—* = 15; f c = 500; /« = 14,000 and M = — — • 
JtL, c 8 

162. A reinforced-concrete floor carrying a combined live and dead load 
of 400 lbs. per sq. ft. is placed over steel beams that are on 6-ft. centers. 

Taking & -is, U-V±, 

allowable compression in concrete 500 lbs. per sq. in., and tension in steel 
14,000 lbs. per sq. in., find depth of slab and area of reinforcement. 



4i8 



GRAPHICS AND STRUCTURAL DESIGN 



163. A rectangular tank supported on its lower edges is 5 ft. wide, 6 ft. 
high and 13 ft. long (inside dimensions). How thick must the floor be made 

to hold the tank full of water ? Allow M = Compression in concrete 

10 

500 lbs. per sq. in.; tension in steel 14,000 lbs. per sq. in. 

Find also the area of the steel reinforcement. 

The common practice in concrete building construction is to use rein- 
forced-concrete beams rather than the steel beams. In this event, owing 
to the monolithic character of the work, part of the floor slab acts as the 
compression flange of the beam. See Chapter XIV. 

164. The T beams have a span of 20 ft., and are spaced 6 ft. center to 
center; see Fig. 59. The combined live and dead load is 250 lbs. per sq. ft. 

Assuming M = ■ ; ~ = 15; fe = 500; f 8 = 14,000; find the thickness 

10 h, c 

of the slab. 



~t 



-6 — 



—60' 



FiG. 59. 




Fig. 60. 



165. Take Problem 164; assume the slab 5 ins. thick and also that d = 
16 ins. for the T beam; find the width of slab required for the flange of the 
T beam and the area of the reinforcing metal. 

Note. — It is usual to limit the width of the slab assumed as flange of the 
T beam. One specification puts this limit at eight times the slab thickness 
but not exceeding one-third the T-beam span. 

Figure 60 shows a plan of a section of a floor. The following data apply 
to all these problems. 

f e - 550 lbs., f, = 14,000 lbs., 



z-» 



Live load 150 lbs. per sq. ft. of floor, dead load 60 lbs. per sq. ft. of floor. 
166. Design the slab, determining the thickness and area of metal. Take 

ur WL 

M = • 

10 



PROBLEMS 419 

167. Assume the slab 5 \ ins. thick, and design the T beam. Assume d = 
19 ins. Find width of flange and area of steel. 

168. Design the girder G, assuming that it carries a T beam B, on each 
side of it, entering it centrally, and that it takes only 85 per cent of the live 
load, i.e., 150 lbs. per sq. ft., but entire dead load of 75 lbs. per sq. ft. As- 
sume the depth of the girder as d = 25 ins. It must be at least deep enough 
to allow its reinforcing bars to pass under the bars in the T beams. Find 
the area of the reinforcement and the width B of the upper flange of the beam. 

Assume the bending as four-fifths that due to a freely supported beam 

centrally loaded, i.e., M= 

169. A beam spans 20 ft. and carries 2500 lbs. per ft. Assume that d = 30 
ins. and find the width and reinforcing area. Take/ 8 = 12,000 lbs., f c = 

„ E a , .- WL 

500 lbs.,— 2 = 15 and M= — — 
E c 8 

170. Take the data given and found for Problem 169, and find the effect 
on f a and f c of placing the steel 2 ins. above the intended position, making 
d = 28 ins. instead of 30 ins. 

171. A slab spans 10 ft. 6 ins. and carries a load of 485 lbs. per sq. ft. 

Assume M = and allow f 3 = 16,000 lbs., f c = 600 lbs. and — s = 15 lbs. 

12 E c 

Find the thickness of the slab and the spacing of f-in. diameter round rods. 

Slabs should always be reinforced in both directions, the metal at right 

angles to the main reinforcement being used to prevent the slab from 

cracking. When the slabs are designed square and reinforced in both 

directions the maximum moment may be assumed as 



172. A square slab reinforced in both directions has a side of 16 ft. and 
carries a combined live and dead load of 220 lbs. per sq. ft. Find thickness 
of slab and area of reinforcement. Use data of Problem 165. 

173. When reinforced-concrete beams are designed assuming that the 

WL 

bending at the center is , what is the bending at the supports? Where 

10 

in the span is the point of zero bending and what should be the minimum 

reinforcing over the supports expressed as a percentage of the reinforcing 

at the center? 

In the problems thus far considered three points have not been taken into 

account. The horizontal shear, the vertical shear and diagonal tension. 

See Chapter XIV. 



420 



GRAPHICS AND STRUCTURAL DESIGN 



174. Derive the following formula for the spacing of stirrups, 

T-jd 

s — • 

V — v c • b *jd 

175. Derive the following formula for the length of rod required on ac- 
count of bonding, 

2U 

176. Derive the following formula for the length of bar required to 
provide flange strength, 



-Vf 



177. Take Problem 158 and determine the vertical reinforcements, allow- 
ing a tensile fiber stress of 16,000 lbs. per sq. in. Neglect the shearing 
strength of the concrete. Try f-in. rounds bent into U stirrups. Would 
stirrups be needed if 50 lbs. per sq. in. was allowed in shear in concrete? 
What about reinforcement at the ends? 

178 to 181. Design a light highway bridge of reinforced concrete. Span 
32 ft. o ins., fill 15 ins., roadway 16 ft. with a 4-ft. wall on each side. 
Assume 5 = 5! ins. See Figs. 61 to 63. 



}-%— 4'o-- 



Fig. 61. 
16 ' » > U-.-±'q!L..^ 



! W' 



7Mttf/M((t(///A 




-5 0" 



-50- 



-5'0- 



4 



-47-0- 



40 Tons on 8 Wneels 

Y 24'0^ 

5^0 "H — Car Width 8'6" 



m& 



ri 



M& 



¥^r 



Fig. 62. 



2W 
3~ 



fc_Q 



f-r.i. 

W=20 Tons 

Fig. 63. 



^ 



178. Determine thickness of slabs under roadway. Assume live load 
500 lbs. per sq. ft., concrete 60 lbs. per sq. ft., fill 120 lbs. per sq. ft. Assume 



PROBLEMS 421 

compressive stress in concrete 650 lbs. per sq. in., tensile stress in steel 
16,000 lbs. per sq. in. Assume that the effect of impact is 50 per cent of 
live-load stresses. 

,, WL , E. 

M = and ■=* = 15. 

10 E c 

Using |-in. round bars, how far apart (center to center) should they be 

WL 

spaced? Since the bending has been taken at M = , what reinforce- 

10 

ments should be placed over girders? 

179. Design girders d and G 6 . Assume dead load of concrete, fill, etc., 

as 900 lbs. per lineal ft. of 32-ft. span. Live load 125 lbs. per sq. ft. of walk 

assumed as 5 ft. o ins. wide. Make b = 12 ins. 

WL 
z = 1.64 corresponding to f c = 650 lbs. M = — — - • 

8 

180. Design girders G z and G 4 on the following basis. Each girder to 
carry one-half the car load or the front wheel of the roller and one-half the 
weight of the rear wheels, Figs. 62 and 63. Assume dead load 1350 lbs. 

per ft. Take M = —— . Add 35 per cent to live-load stresses to allow for 
8 

impact. How many i|-in. bars will be required? Assume d = 34.5, 

/, = 16,000 lbs. What width of the flange will be stressed? Make stem 

14 ins. wide. 

181. Draw diagram of maximum shears covering dead load and trolley 
load. Assume ^-in. U stirrups and find the spacing at the abutment and 
at 2, 4, 6 and 8 ft. from this end, allowing 60 lbs. per sq. in. shear on the 
concrete and using the formula of Problem 174. 

182. In a given beam d = 30 ins., four i|-in. round reinforcing bars run 
parallel with the lower flange through the entire span. The maximum 
reaction is 44,000 lbs. What is the bond stress per square inch and is this 
satisfactory if the specifications allow 60 lbs. per sq. in. ? How could this 
be provided for ? 

183. In a beam d = 34 ins., three i-in. bars run parallel with the 
lower flange through the entire span. The maximum reaction is 19,000 lbs. 
What is the bond stress per square inch and is this satisfactory if the speci- 
fications allow a bond stress of 60 lbs. per sq. in. ? If not satisfactory, how 
could it be provided for ? 



422 



GRAPHICS AND STRUCTURAL DESIGN 



Reinforced-concrete Columns 

184. Allowing 600 lbs. per sq. in. on the concrete, what load can be placed 
on a column 11 ft. high, 12 X 12 ins. and reinforced by four §-in. bars? 
When a column's height does not exceed twelve times its least dimension 
the influence of length upon its buckling can be neglected. Assume column 
dimensions inside bars as 10 X 10 ins. 

185. Load 420,000 lbs., reinforcement 1 per cent, height 15 ft. What 
size square columns allowing 600 lbs. upon concrete will be required? How 
many i-in. bars will be required? What would you make the outside 
dimensions of the column? 

186. Take the truss in Fig. 64, place a load L at the point indicated, 
show the necessity for the double lacing. Explain why this is usual in the 
central panels and frequently not necessary at the end panels. 





Fig. 64. 



Fig. 65. 



187. With Fig. 65, by means of force and equilibrium polygons find the 
required depth of a concrete foundation 8 ft. square, at 125 lbs. per cu. ft. so 
that the moment of the foundation about the toe A shall be twice that of the 
crane weight and live load about the same point. At the bottom of the 
foundation make a diagram showing the distribution of pressure on the soil. 

188. Take the data given and found for Problem 187; turn the crane until 
the load lies in the line of a diagonal of the foundation. With the resultant 
pressure upon the soil as found in Problem 187, determine the maximum and 
minimum pressure on the foundation by the method given in Chapter XVI. 

189. Take Fig. 66, and by means of a force and an equilibrium polygon 
determine the maximum load the locomotive crane can lift in the position 
given. Boom is 36 ft. long, and weighs 2000 lbs., it is attached to cab 5 ft. 
from center line of track and is shown at 30 degrees with the horizontal. 
Gauge of track is 4 ft. 8| ins. Counterweight is 15,000 lbs. acting at center 
line of track. Engine, machinery, etc., weigh 18,000 lbs., and act 12 ins. to 
left of center line of track. Boiler and base weigh 12,000 lbs., and act 7 ft. 
6 ins. to left of center line of track. Cab, etc., weigh 16,000 lbs., and act at 
center line of track. 



PROBLEMS 



423 



190. Design a concrete footing, Fig. 67, to carry 200 tons from the column 

through a shoe 4 ft. square, resting on reinforced concrete. The soil can 

carry 3 tons per sq. ft. What spacing of f-in. round bars will be required? 

Assume 

E a 

E c 



fc= 55o; /*= 16,000 lbs.; 



IS- 




Fig. 66. 



Fig. 67. 



What advantages would this type of foundation have over the usual con- 
crete pier? Consult Chapter XV. 

191. In Fig. 68 assume sag 60 per cent of width, coal (bituminous) 50 
lbs. per cu. ft. Area of surcharged bin A' = 0.57 a 2 . Assume that the 
load on the bin varies from the sides to the middle as the intercepts between 




% Plate 

W'Rivets 



Plate 34 Plate J % Plate 96 Plate j 
Etivets %"Rivets | ^"Rivets %'ltivets j . 

-30- — >+< 80- >r* 30" >h 30^— H ' 



r 



Fig. 68. 



-30 
-120 0- 

Fig. 69. 



the sides of a triangle. Show by means of force and equilibrium polygons 
how to find the tension in the bottom of the bin plate and also in the sides 
at the point of suspension from the side girders. 

192. A stack, Fig. 69, has outside diameter of 7 ft. and is 120 ft. high. 
Assume a uniform wind pressure of 30 lbs. per sq. ft. acting on it. The steel 



424 GRAPHICS AND STRUCTURAL DESIGN 

weighs 30,000 lbs. and the concrete base is 20 it. square. If the concrete 
weighs 125 lbs. per cu. ft., how deep must the foundation be if the resultant 
pressure falls within the kern and what is the distribution of pressure on 
the soil? 

193. The 7-ft. stack shown in Fig. 69 is 1 20 ft. high, is made of -^-in. plate 
for 30 ft. at the top, and each succeeding 30 ft. toward the foot of the stack 
is made T \ in. thicker. What is the extreme fiber stress in the net section, 
60 ft. from the top, assuming 20 per cent of the plate cut out for rivets? 
Calculate the pitch of rivets, allowing 8000 lbs. in shear and 16,000 lbs. in 
bearing and assume rivets f in. in diameter. 

194. In Fig. 69 investigate the fiber stress and rivet pitch for the seam 
80 ft. from the top. 

195. A stack is 125 ft. high. Its outside diameter is 7 ft. 6 ins. If 
12 bolts are used on a circle whose radius is 65 ins., what must their 
diameter be? Allowable tension is 12,000 lbs. per sq. in. Weight of stack 
is 30,000 lbs. 



ANSWERS TO PROBLEMS 



5. 


0.79 in. 


6. 


2.01 ins. 


7. 


3.82 ins. 


8. 


3.86 ins. 


14. 


13-5- 


15. 


3.90. 


16. 


I33-4- 


17. 


14.6. 


18. 


324.0. 


19. 


241. 1. 


20. 


0.0069 d 4 . 


ax. 


11,340.1. 


24. 


r = w,M = — and. 
3 




_ W % , WL 


27. 


K = , Mmax — — 7~- 
2 




Wx 2 Wx* 




Mx= ■ Fi" 

2 3 U 



Wx 3 
3#" 



w 

28. Ri = — , A'max = O.578 Z,, 
3 

Mmax = O.129 WL. 
30. i?i = 7000 lbs., 

Mmax = 78,000 ft. lbs., 



45. 



46. 
47- 
48. 
49. 
50. 
51. 

53- 
54- 
55. 
58. 
59. 
60. 
61. 
63. 



Load 19,200 lbs., fiber stress in fixed 
beam 10,670 lbs. per sq. in., de- 
flection in supported beam 0.534 
in., deflection in fixed beam 0.107 
in. 

2\ ins. diam. 

15,500 lbs. 

4.92 ins. back to back, r = 3.10. 

r = 1.91, n = 1.73. 

r = 1.00, n = 2.53. 

r = 2.0, r = 2.5. 

6150, 7600 and 6550 lbs. 

11,800, 11,750 and 11,750 lbs. 

203,000 lbs. 

34,880 lbs. 

64,400 lbs. 

66,300 lbs. 

C = 16,740 lbs.; P = 19,100 lbs. 

One 12-in. channel at 20I lbs. hori- 
zontally and on top of one 20-in. 
I beam at 65 lbs. placed verti- 
cally. Lateral stress 5700 lbs. 
per sq. in. 





Mn = 75,000 ft. lbs. 




66. 


59.4 ins. 


32- 


Mmax = 446,490 ft. lbs., 




67. 


5760 lbs. 




if 28 = 443,600 ft. lbs. 




68. 


6400 lbs. 


34. 


15-in. I at 45 lbs. 




69. 


16 ft. 8 ins. 


35. 


15-in. I at 50 lbs. 




70. 


4740 lbs.; A = 0.486 i] 


36. 


-required 193.2 use two 


20-in. I 


7i- 
72. 


1420 lbs. 
10,940 lbs. 




beams at 65 lbs. 




73- 


8125 lbs. 


37. 


15-in. I at 42 lbs. 




74. 


8450 lbs. compression. 


38. 


18-in. or 20-in. I at 65 lbs. 




75. 


11,660 lbs. compression, 


39. 


20-in. I at 65 lbs. 




76. 


7 rivets. 


40. 


- = 123.8. 20-in. I at 75 


lbs. 


77- 


12 rivets. 




e 




78. 


10 rivets. 


43. 


2i r ooo lbs. 




79. 


81,000 lbs. 


44. 


Load 1460 lbs., deflection 0.161 in. 


80. 


57,600 lbs. 



425 



426 



GRAPHICS AND STRUCTURAL DESIGN 



81. 52,500 lbs. 

83. 21.8 sq. ins. 

84. 24.9 sq. ins. 

85. 21.7 sq. ins. 

86. Top flange two 4 X 4 X §*-in. angles, 

and one plate 22 X ig-in., lower 
flange two 4 X 4 X f-i n - angles 
and one plate 22 X ^-in. 

87. 10,990 in. lbs., and 916 lbs. 

88. 13,400 in. lbs. and 1117 lbs. 

89. The twisting moment of a round 

shaft may be 1.2 times that on a 
square shaft of the same weight 
per foot. 

pd 3 N 



90. H.P. 



321,300 



AG 



DH 



91. 2I. 3 U . 

92. 2 ^ ins. 

93. 2§ ins. 

94. GC = 37,600 lbs. comp. 

33,600 lbs. tension. 

95. GH = 4300 lbs. comp.. 

35,460 lbs. comp. 

96. HI = 4800 lbs. tension and AI = 

28,800 lbs. tension. 

97. FL = 31,180 lbs. comp. and LM = 

14,400 lbs. tension. 

98. BF = 154,660 lbs. comp., FA = 

109,375 lbs. tension. 

99. FG = 110,470 lbs. tension, GC = 

187,500 lbs. compression. 

100. GH = 31,250 lbs. comp., — HI = 

66,280 lbs. comp. 
114. A J = 166,670 lbs. comp., KJ = 

25,000 lbs. tens., IF = 104,100 

lbs. tens. 
129. 18 rivets. 



I 3°« 3, the spacing is commonly not 

allowed to exceed 4 to 6 ins. 
^S* 4-6 ins. 
136. 4.8 ins. 
137- 3-5 ins. 

138. 2.28 ins. 

139. 2.8 ins. 

140. 2.93 ins. 

141. 4.38 ins. 

142. 14.67 ins. 

143. 9.6 ins. 

144. 0.27 in. 

145. 2 rows. 

146. Two plates each f in. thick. 

147. 2 rows. 

148. 6 rivets. 

149. Check by method of coefficients. 

158. b = 13.5 ins. and A = 1.34 sq. ins. 

159. d = 24 ins., b = 9.5 ins., A = 
- 1. 14 sq. ins. 

161. d = 2.2 ins. and A = 0.166 sq. in. 

162. d = 4.9 ins. and A = 0.372 sq. in. 
169. b = 20 ins. and A = 4.8 sq. ins. 
170- fc = 555 lbs. per sq. in. and f 3 = 

12,800 lbs. 
171. d = 6.85 ins. Spacing = 6.77 ins. 

178. d = 4.65 ins., |-in. bars 5.5-in. 

centers. 

179. d = 38.0 ins., five — xib m - 0bars. 

180. Eight i|-in. bars or nine if-in. 

bars. 

182. 106 lbs. per sq. in. 

183. 68 lbs. per sq. in. 

184. 66,590 lbs. 

185. 25 ins. square and eight i-in. 

bars. 



INDEX 



Algebraic determination of stresses, 52. 
Angles of repose and weights of materials, 286. 
Arch floors, brick and tile, 329. 
Ash bins, pressures on, 299. 

Base for reinforced-concrete chimney, 279. 

Beams, design of, 76. 

Beams of uniform section, bending moments, deflections, etc., 7. 

Beams, reinforced-concrete : 

approximate formulae for, 201. 

bent bars, 224. 

bond stresses, 220. 

design of a T beam, 223. 

diagonal tension in, 216. 

forms of reinforcements, 231. 

horizontal shear in, 215. 

parabolic variation of stresses, 214. 

reinforcing rods, lengths of, 221. 

stirrup spacing in, 218. 

theory of beams with double reinforcement, 210. 
rectangular beams, 199. 
slabs, 209. 
T beams, 204. 
Beams unsupported laterally, 100. 
Bearing and shearing value of rivets, 85. 
Bed plates for plate girders for railway bridge, 162. 
Bending due to moving loads, 47. 
Bending moments on beams, 77. 

Bending moments, deflections, etc., for beams of uniform section, 7. 
Bending moments on a railway girder due to locomotive and train load, 144. 
Bents, 114. 

Bent bars in T beams, 224. 
Bethlehem rolled sections, 18. 
Bins: 

coefficients of friction between materials, 305. 

design of, 315. 

graphical determination of forces acting on, 305. 

427 



428 INDEX 

Bins (continued). 

pressure on grain bins, 298. 

shallow bins, 299. 

stresses in bins, formulae for, 299. 
Bond stresses between concrete and steel, 220. 
Box girder for electric overhead traveling crane, 178. 
Bracing for steel mill buildings, 120. 
Bricks, 4. 
Bricks, fire, 4. 
Brick arch floors, 328. 
Brick chimney, design of, 251. 
Brick floors, 325. 
Brick walls, 332. 
Buildings, foundations for, 238. 
Bunkers, suspension, 309. 

Cast iron, 1. 

Cement, 195. 

Cement concrete floors, $2$. 

Center of gravity, graphical determination of, 30. 

Centers of gravity of angles, 15. 

channels, 16. 
Character of stresses, 43. 
Chimney, brick: 

design of a, 251. 

kern of a section, 247. 
Chimney, reinf orced-concrete : 

base, 279. 

design of a reinforced-concrete, 268. 

reinforcements, 276. 
Chimney, steel: 

design of a, 257. 

foundation, 262. 

foundation bolts, 266. 

lining, 262. 

maximum pressure on the soil, 264. 

ring seams, 260. 

thickness of shell, 257. 

vertical seams, 260. 
Clay tile roofing, 350. 
Clearance for cranes, 345. 
Coal bins, pressures on, 299. 
Column formulae, 91. 
Columns for steel mill buildings, 132. 
Columns, 91. 
Columns, theory of reinforced-concrete, 229. 



INDEX 429 



Columns, timber, 4. 

Combined stresses, 97. 

Compression pieces, design of, 76. 

Concrete (see also Reinforced concrete). 

Concrete blocks, hollow, 334. 

Concrete retaining wall, design of, 288. 

Concrete roofing, 350. 

Concrete walls, solid, 333. 

Concrete wearing surfaces, 330. 

Concurrent and nonconcurrent forces, 21. 

Continuous beams, 32. 

Conveyors, girders for, 103. 

Cooper's E-60 loading, moment table for, 72. 

Coplanar and noncoplanar forces, 21. 

Corrugated bars, 232. 

Corrugated steel roofing, 346. 

Corrugated steel sides, 332. 

Couple, 21. 

Cranes: 

electric overhead traveling crane, 177. 

top-braced jib crane, 171. 

underbraced jib crane, 165. 
Crane clearances, 345. 
Crane frames (see Cranes). 

Deflection of beams, graphical determination of, 30. 
Deflection, bending moments, etc., 7. 
Design of beams, 76. 

compression pieces, 76. 
frame for top-braced jib crane, 171. 
frame for underbraced jib crane, 165. 
railway girder, 144. 
roofs, 346. 

roof truss members, 127. 
steel mill building, No. 1, 120. 
steel mill building, No. 2, 137. 
tension pieces, 76. 
Determination of stresses, algebraic, 52. 

method of coefficients, 55. 
method of moments, 54. 
Diagonal tension in reinforced concrete beams, 216. 
Diagram of maximum live-load shears, 49. 
Dimensions of angles, 14. 

Bethlehem beams, 19. 
channels, 16. 
edged plates, 13. 



430 INDEX 

Dimensions (continued). 

girder beams, 19. 

I beams, 17. 

H columns, 18. 

sheared plates, 12. 
Direction of a force, 20. 

Elastic limit of concrete, 196. 
Electric overhead traveling crane: 
design of box girder, 178. 
flange areas, 179. 
rivets, 184. 
girder with channel flanges, 185. 
girder with horizontal stiffening girder, 188. 
stiffeners, 185. 
specifications for, 177. 
Equilibrant, 21. 
Equilibrium, 21. 

. polygon, 23. 
Expanded metal, 232. 

Fiber stresses in underbraced jib cranes, 167. 

working, 3. 
Flange areas for girders of E.O.T. cranes, 179. 

of plate girders, 149. 
Flange plates of plate girders for railway bridges, lengths of, 151. 
Flange rivets in girder for E.O.T. crane, 184. 
Flange rivets in plate girder for railway bridge, 152. 
Flat plates, strength of, 9. 
Floors, ground: 

brick, 325. 

cement concrete, 323. 

wooden, 326. 

wooden block, 325. 
Floors, upper: 

brick arch, 328. 

concrete wearing surfaces, 330. 

hollow tile arch, 329. 

iron floors, 331. 

reinforced-concrete, 330. 

steel, 328. 

steps, 331. 

wooden, 327. 
Floor-beam reaction, maximum, 69. 
Force, 20. 
Force and equilibrium polygons, uses of, 28. 



INDEX 431 

Force polygon, 22. 

Force triangle, 21. 

Forces acting in bins, graphical determination of, 305. 

Formulae for piles, 246. 

for stresses in bins, 300. 
Foundation bolts for chimneys, 266. 
Foundations for buildings, 238. 

for machinery, 233. 

piles, 244. 

pile formulas, 246. 

pressure on soil, allowable, 236. 
Foundations for chimneys, 262. 
Framing, standard, 80. 
Friction, coefficient of, between materials, 305. 

Girders for conveyors, 103. 

Girders for E.O.T. crane, specifications for, 177. 

with channel flanges for E.O.T. cranes, 185. 

with horizontal stiffening girders for E.O.T. cranes, 188. 
Glass, 346. 

Grain bins, pressures on, 298. 
Graphic moments, 26. 
Graphics, 20. 

statics, 20. 
Graphical determination of forces acting in bins, 305. 

of stirrup spacing in reinforced-concrete beams, 219. 
Gravel, 196. 

Gravel and slag roofing, 350. 
Ground floors, 326. 

Hollow tile arch floors, 329. 

Horizontal shear in reinforced-concrete beams, 215. 

stiffening girders for E.O.T. crane girders, 189. 

Impact, 146. 

Inertias of geometrical sections, 10. 

angles, 14. 

channels, 16. 

H columns, 18. 

I beams, 17. 
Influence diagrams, 61. 
Iron, cast, 1. 
Iron floors, 331. 

Jib for underbraced crane, selection of, 167. 
Jib crane, underbraced, 165. 
top-braced, 171. 



43 2 INDEX 

Kahn trussed bar, 232. 
Kern of a section, 247. 

Lateral bracing for plate-girder railway bridge, 156. 

Line of action, 20. 

Lining, chimney, 262. 

Live-load shears, diagram of maximum, 49. 

I stresses, 56. 

Long beams unsupported laterally, 100. 

Machinery, foundations for, 233. 

Magnitude of a force, 20. 

Masonry retaining walls, design of, 288. 

Mast of a top-braced jib crane, 176. 

Mast of an under-braced jib crane, 169. 

Materials, 1. 

Maximum bending due to moving loads, 47. 

floor-beam reaction, 69. 

live-load shears, diagram of, 49. 
. moment due to moving loads, 61. 

shear due to moving loads, 64. 
Members of frame of top-braced jib crane, 173. 

of underbraced jib crane, 169. 
Metals, physical properties of, 2. 
Method of coefficients for the determination of stresses, 55. 

moments, determination of stresses by, 54. 
Mixing concrete, 196, 
Moment of a couple, 21. 

table for Cooper's E-60 loading, 72. 
Moving loads carried under trusses, 2>1- 

maximum bending due to, 47. 

Physical properties of concrete, 196. 

metals, 2. 
Pile formulae, 246. 
Piles, 244. 
Plate girder for railway bridge: 

bed plates, 162. 

flange rivets, 152. 

flange plates, lengths of, 151. 

rivet spacing in flange angles, 154. 

stiffening angles, 151. 

web splice, 157. 

wind bracing, 156. 
Plate girders: 

dead-load shear, 149. 

maximum end shear, 147. 



INDEX 433 



Plate-girder railway bridge, 144. 
Plates, strength of flat, 9. 

dimensions of edged, 13. 
of sheared, 12. 
Point of application of a force, 20. 
Pressures on retaining walls, 282. 
soils, allowable, 236. 
soil under chimney, maximum, 264. 
Properties of concrete, physical, 197. 
metals, 2. 
sections, 10. 
Properties of timber, 3. 

table of, 5. 
Purlins, 123. 



Railway girder, design of, 144. 

girders, impact allowance for, 146. 
Ransome twisted bars, 232. 
Rectangular beams, theory of, 199. 
Reinforcing frame, 232. 

rods, lengths of, 221. 
Reinforced concrete: 

approximate formulae for rectangular beams, 201. 

bond stresses, 220. 

cement for, 195. 

columns, 228. 

design of T beams, 223. 

diagonal tension, 217. 

elastic limit, 196. 

forms of reinforcements, 231. 

gravel, 196. 

horizontal shear in beams, 215. 

mixing, 196. 

parabolic variation of stress, 214. 

physical properties, 196. 

reinforcing rods, lengths of, 221 

sand, 195. 

stone, 196. 

stirrup spacing in beams, 218. 

theory of beams with double reinforcement, 210. 
rectangular beams, 199. 
slabs, 209. 
T beams, 204. 
Reinforced-concrete chimney, 268. 

retaining walls, 289. 



434 INDEX 

Reinforced- concrete walls, 334. 
Reinforcements, forms of, 232. 
Resistances of sections, 10. 
Resultant, 21. 
Retaining walls: 

design of a masonry, 288. 

reinforced-concrete, 289. 

distribution of pressure on, 284. 

graphical determination of the pressures on, 283. 

pressures on, 282. 

weights and angles of repose of materials in fills, 286. 
Ring seams of steel chimneys, 260. 
Rivet spacing in flange angles, 152. 
Rivet values in shear and bearing, table of, 85. 
Riveting, 81. 
Riveting, rules for, 89. 
Roofs, types of, 123. 
Roof coverings: 

clay tile, 350. 

concrete, 350. 

corrugated steel, 346. 

slag or gravel, 350. 

slate, 349. 



Sand, 195. 

Sections, inertias, resistances, etc., 10. 

properties of, 10. 
Shallow bins, pressures on, 299. 
Shear and bearing values of rivets, 85. 
Shear in reinforced-concrete beams, horizontal, 215. 
Shell thickness of steel chimney, 257. 
Shop floors (see Floors). 
Slag or gravel roofing, 350. 
Slate roofing, 349. 
Soil, allowable pressure on, 236. 
Specifications for girders of an E.O.T. crane, 177. 
Splice for plate-girder of railway bridge, web, 157. 
Standard framing, 79. 
Steel castings, 2. 
Steel floors, 328. 
Steel mill buildings: 

design of Problem 1, 120. 

design of Problem 2, 137. 
Steps, 331. 
Stiffeners for girders for E.O.T. crane, 185. 



INDEX 435 



Stiff eners for plate girders of railway bridge, 151. 
Stiffening steel mill buildings, 120. 
Stirrup spacing in reinforced-concrete beams, 218. 
Stone, 196. 

walls, 333. 
Strength of flat plates, 9. 
Stresses : 

algebraic determination of, 52. 

character of, 43. 
Stresses in concrete, unit working, 197. 

structures, 35. 
Structural material, 11. 
Structures, stresses in, 35. 

T beams: 

bent bars, 224. 

design of a, 223. 

web reinforcements, 225. 
Tension pieces, design of, 76. 
Tension or compression in a member, 36. 
Thatcher bar, 232. 
Theory of columns, 229. 
Tile arch floors, 329. 
Timber columns, 4. 
Timber, properties of, 3. 

table of properties of, 5. 
Top-braced jib crane: 

design of frame, 171. 

mast, 176. 

selection of members, 173. 
Towers, 114. 

Traveling crane (see Electric overhead traveling crane), 177. 
Triangular mesh steel reinforcement, 232. 
Truss, determination of the stresses using moment table, 74. 
Trusses, 112. 
Trusses carrying moving loads under them, 37. 

wind loads on, 39. 
Twisted bars, Ransome, 232. 

Underbraced jib crane: 

design of frame, 165. 

fiber stresses in, 167. 

mast, 169. 

members of frame, 169. 

selection of jib, 167. 
Unit working stresses for concrete, 197. 



436 INDEX 

Vertical seams of a steel chimney, 260. 
shear in beams, 77. 



Walls: 

brick, 332. 

corrugated steel, 332. 

glass, 346. 

hollow concrete blocks, 334. 

reinforced-concrete, 334. 

solid concrete, $$^. 

stone, S33- 

wooden, 332. 
Wearing surfaces, 330. 
Web reinforcements in T beams, 225. 
Web splice for plate girders of railway bridge, 157. 
Weights and angles of repose of materials, 286. 
Wind bracing for girders carrying conveyors, 109. 
plate-girder railway bridge, 156. 
steel mill building, 134. 
Wind load on trusses, 39. 
Wooden block floors, 325. 
Wooden floors, 327. 

walls or sides, 332. 
Working fiber stresses, 3. 



